| PROJECT | : MAA BHARTI NURSING COLLEGE, GAURAVPATH, KUSHALGARH |
| PLAN | : nurse |
| JOB NO. | : 1 | REF. NO. | : |
| DATE | : 11/06/25 | TIME | : 09:43:38 |
| Building Version | : 1.640 | STRUDS Version | : 4.0.0 |
Terms Used In Calculation :
In biaxial column design,
Pu = Puc + Pus(Total)| where, |
| Pu | = | external axial compressive load, |
| Puc | = | axial compressive resistance offered by concrete, |
| Pus(Total) | = | total axial compressive resistance offered by steel |
| | | at different levels in the section. |
Mu = Muc + Mus(Total)| where, |
| Mu | = | external moment about centroidal axis, |
| Muc | = | moment of resistance offered by concrete in compression, |
| Mus(Total) | = | total moment of resistance offered by steel |
| | | at different levels in the section. |
| i | = | serial no. of row of reinforcement, |
| Asi | = | cross-sectional area of reinforcement in the i th row, |
| fsi | = | stress in the reinforcement in the i th row (compression + ve, tension - ve), |
| fci | = | compressive stress in concrete at the level of i th row of reinforcement, |
| ei | = | strain at the i th row of reinforcement from the stress-strain curve of steel and concrete, |
| xi | = | distance of the bars in the i th row from the centroid of the section |
Values of stress in steel :
Design strength in bending compression (fyd) = 0.87 fy
(a) Fe 250,
For ei >= fyd / Es, fsi = fyd
For ei < fyd / Es, fsi = ei x fyd
(b) Fe 415,
For ei >= 0.8 x fyd / Es, fsi = value obtained from stress-strain curve
For ei < 0.8 x fyd / Es, fsi = ei x fyd| Stress | Stress level
(N/mm˛) | Strain |
| 0.800 fyd | 288.7 | 0.00144 |
| 0.850 fyd | 306.7 | 0.00163 |
| 0.900 fyd | 324.8 | 0.00192 |
| 0.950 fyd | 342.8 | 0.00241 |
| 0.975 fyd | 351.8 | 0.00276 |
| 1.000 fyd | 360.9 | 0.00380 |
Values of stress in concrete :
For ei >= 0.002, fci = 0.446 fck
For ei < 0.002, fci = (446.ei x (1 - 250.ei)) x fck
Values of strain :
Values of strain at different levels are obtained by taking maximum strain
as 0.0035 as the reference value at the highly compressed edge.
Design of CG1 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 199.50 kN |
| MomentX,(Mx) | = | 37.38 kN-m |
| MomentY,(My) | = | 25.27 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C9 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 4.130 kN.m |
| My_MinEccen | = | 3.990 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 37.384 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 25.273 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 37.38/199.50 = 187 mm |
| Actual eccenY | = | My / P = 25.27/199.50 = 127 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(187,21) = 187 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(127,20) = 127 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(187 mm) > 0.05 x 300(15 mm)
and eccenY(127 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00247 | 344.22 | 8.92 | 335.30 | 75.84 | 179 | 13.58 |
| 2 | 226 | 0.00045 | 90.29 | 3.57 | 86.72 | 19.62 | 90 | 1.76 |
| 3 | 226 | -0.00156 | -300.27 | 0.00 | -300.27 | -67.92 | 0 | 0.00 |
| 4 | 226 | -0.00358 | -358.94 | 0.00 | -358.94 | -81.19 | -90 | 7.27 |
| 5 | 226 | -0.00559 | -360.90 | 0.00 | -360.90 | -81.63 | -179 | 14.61 |
| Total | -135.29 | | 37.21 |
| xux | = | 156 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 156/450 = 0.124 |
| Puc | = | 0.124 x 20.00 x 300 x 450 |
| = | 336.02 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 336.02 + (-135.29) | = | 200.74 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 156/450 |
| = | 0.144 |
| Muc | = | 336.02 x (0.5 x 450 - 0.144 x 450) = 53.86 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 53.86 + (37.21) |
| = | 91.07 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00141 | 282.09 | 8.14 | 273.95 | 154.91 | 104 | 16.11 |
| 2 | 565 | -0.00804 | -360.90 | 0.00 | -360.90 | -204.08 | -104 | 21.22 |
| Total | -49.17 | | 37.34 |
| xuy | = | 77 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 77/300 = 0.092 |
| Puc | = | 0.092 x 20.00 x 450 x 300 |
| = | 249.64 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 249.64 + (-49.17) | = | 200.47 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 77/300 |
| = | 0.107 |
| Muc | = | 249.64 x (0.5 x 300 - 0.107 x 300) = 29.44 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 29.44 + (37.34) |
| = | 66.78 kN-m |
| Pu/Puz | = | 0.128 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.128, | an | = | 1.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((37.38/91.07)1.000) + ((25.27/66.78)1.000) |
| = 0.789 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG1 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 419.82 kN |
| MomentX,(Mx) | = | 32.28 kN-m |
| MomentY,(My) | = | 23.17 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C9 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 8.690 kN.m |
| My_MinEccen | = | 8.396 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 32.275 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 23.170 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 32.28/419.82 = 77 mm |
| Actual eccenY | = | My / P = 23.17/419.82 = 55 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(77,21) = 77 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(55,20) = 55 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(77 mm) > 0.05 x 300(15 mm)
and eccenY(55 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00273 | 351.12 | 8.92 | 342.20 | 77.40 | 179 | 13.86 |
| 2 | 226 | 0.00124 | 248.46 | 7.64 | 240.82 | 54.47 | 90 | 4.88 |
| 3 | 226 | -0.00025 | -49.79 | 0.00 | -49.79 | -11.26 | 0 | 0.00 |
| 4 | 226 | -0.00174 | -313.58 | 0.00 | -313.58 | -70.93 | -90 | 6.35 |
| 5 | 226 | -0.00323 | -355.93 | 0.00 | -355.93 | -80.51 | -179 | 14.41 |
| Total | -30.82 | | 39.49 |
| xux | = | 210 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 210/450 = 0.168 |
| Puc | = | 0.168 x 20.00 x 300 x 450 |
| = | 453.73 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 453.73 + (-30.82) | = | 422.90 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 210/450 |
| = | 0.194 |
| Muc | = | 453.73 x (0.5 x 450 - 0.194 x 450) = 62.44 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 62.44 + (39.49) |
| = | 101.93 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00231 | 338.96 | 8.92 | 330.04 | 186.63 | 104 | 19.41 |
| 2 | 565 | -0.00310 | -354.75 | 0.00 | -354.75 | -200.60 | -104 | 20.86 |
| Total | -13.97 | | 40.27 |
| xuy | = | 135 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 135/300 = 0.162 |
| Puc | = | 0.162 x 20.00 x 450 x 300 |
| = | 436.64 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 436.64 + (-13.97) | = | 422.67 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 135/300 |
| = | 0.187 |
| Muc | = | 436.64 x (0.5 x 300 - 0.187 x 300) = 41.02 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 41.02 + (40.27) |
| = | 81.29 kN-m |
| Pu/Puz | = | 0.270 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.270, | an | = | 1.116 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((32.28/101.93)1.116) + ((23.17/81.29)1.116) |
| = 0.523 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG1 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 628.56 kN |
| MomentX,(Mx) | = | 29.11 kN-m |
| MomentY,(My) | = | 20.90 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C9 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 13.011 kN.m |
| My_MinEccen | = | 12.571 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 29.108 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 20.904 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 29.11/628.56 = 46 mm |
| Actual eccenY | = | My / P = 20.90/628.56 = 33 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(46,21) = 46 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(33,20) = 33 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(46 mm) > 0.05 x 300(15 mm)
and eccenY(33 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00289 | 352.91 | 8.92 | 343.99 | 77.81 | 179 | 13.93 |
| 2 | 226 | 0.00170 | 310.78 | 8.71 | 302.07 | 68.33 | 90 | 6.12 |
| 3 | 226 | 0.00050 | 100.67 | 3.92 | 96.74 | 21.88 | 0 | 0.00 |
| 4 | 226 | -0.00069 | -137.73 | 0.00 | -137.73 | -31.15 | -90 | 2.79 |
| 5 | 226 | -0.00188 | -322.34 | 0.00 | -322.34 | -72.91 | -179 | 13.05 |
| Total | 63.95 | | 35.88 |
| xux | = | 263 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 263/450 = 0.210 |
| Puc | = | 0.210 x 20.00 x 300 x 450 |
| = | 567.63 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 567.63 + (63.95) | = | 631.58 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 263/450 |
| = | 0.243 |
| Muc | = | 567.63 x (0.5 x 450 - 0.243 x 450) = 65.66 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 65.66 + (35.88) |
| = | 101.55 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00262 | 348.18 | 8.92 | 339.26 | 191.85 | 104 | 19.95 |
| 2 | 565 | -0.00136 | -272.58 | 0.00 | -272.58 | -154.14 | -104 | 16.03 |
| Total | 37.71 | | 35.98 |
| xuy | = | 183 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 183/300 = 0.219 |
| Puc | = | 0.219 x 20.00 x 450 x 300 |
| = | 592.31 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 592.31 + (37.71) | = | 630.02 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 183/300 |
| = | 0.254 |
| Muc | = | 592.31 x (0.5 x 300 - 0.254 x 300) = 43.80 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 43.80 + (35.98) |
| = | 79.78 kN-m |
| Pu/Puz | = | 0.404 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.404, | an | = | 1.340 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((29.11/101.55)1.340) + ((20.90/79.78)1.340) |
| = 0.354 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG1 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 819.99 kN |
| MomentX,(Mx) | = | 24.11 kN-m |
| MomentY,(My) | = | 17.12 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C9 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 16.974 kN.m |
| My_MinEccen | = | 16.400 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 24.106 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 17.119 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 24.11/819.99 = 29 mm |
| Actual eccenY | = | My / P = 17.12/819.99 = 21 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(29,21) = 29 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(21,20) = 21 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(29 mm) > 0.05 x 300(15 mm)
and eccenY(21 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00299 | 353.77 | 8.92 | 344.85 | 78.00 | 179 | 13.96 |
| 2 | 226 | 0.00198 | 327.16 | 8.92 | 318.24 | 71.98 | 90 | 6.44 |
| 3 | 226 | 0.00098 | 196.63 | 6.61 | 190.02 | 42.98 | 0 | 0.00 |
| 4 | 226 | -0.00002 | -3.60 | 0.00 | -3.60 | -0.81 | -90 | 0.07 |
| 5 | 226 | -0.00102 | -203.83 | 0.00 | -203.83 | -46.11 | -179 | 8.25 |
| Total | 146.05 | | 28.73 |
| xux | = | 313 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 313/450 = 0.250 |
| Puc | = | 0.250 x 20.00 x 300 x 450 |
| = | 675.84 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 675.84 + (146.05) | = | 821.89 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 313/450 |
| = | 0.289 |
| Muc | = | 675.84 x (0.5 x 450 - 0.289 x 450) = 64.10 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 64.10 + (28.73) |
| = | 92.83 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00275 | 351.63 | 8.92 | 342.71 | 193.80 | 104 | 20.15 |
| 2 | 565 | -0.00062 | -124.58 | 0.00 | -124.58 | -70.45 | -104 | 7.33 |
| Total | 123.35 | | 27.48 |
| xuy | = | 216 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 216/300 = 0.259 |
| Puc | = | 0.259 x 20.00 x 450 x 300 |
| = | 698.63 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 698.63 + (123.35) | = | 821.97 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 216/300 |
| = | 0.299 |
| Muc | = | 698.63 x (0.5 x 300 - 0.299 x 300) = 42.13 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 42.13 + (27.48) |
| = | 69.61 kN-m |
| Pu/Puz | = | 0.527 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.527, | an | = | 1.545 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((24.11/92.83)1.545) + ((17.12/69.61)1.545) |
| = 0.239 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG1 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3375 mm |
From analysis results,loads on column
| Axial load,(P) | = | 983.05 kN |
| MomentX,(Mx) | = | 14.03 kN-m |
| MomentY,(My) | = | 10.07 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C9 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 21.381 kN.m |
| My_MinEccen | = | 19.661 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 21.381 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 19.661 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 14.03/983.05 = 14 mm |
| Actual eccenY | = | My / P = 10.07/983.05 = 10 mm |
| eccenXMin | = | (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(14,22) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(10,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00305 | 354.37 | 8.92 | 345.45 | 78.14 | 179 | 13.99 |
| 2 | 226 | 0.00218 | 334.50 | 8.92 | 325.58 | 73.64 | 90 | 6.59 |
| 3 | 226 | 0.00131 | 262.93 | 7.87 | 255.05 | 57.69 | 0 | 0.00 |
| 4 | 226 | 0.00045 | 89.07 | 3.53 | 85.54 | 19.35 | -90 | -1.73 |
| 5 | 226 | -0.00042 | -84.79 | 0.00 | -84.79 | -19.18 | -179 | 3.43 |
| Total | 209.64 | | 22.28 |
| xux | = | 360 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 360/450 = 0.288 |
| Puc | = | 0.288 x 20.00 x 300 x 450 |
| = | 778.36 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 778.36 + (209.64) | = | 988.00 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 360/450 |
| = | 0.333 |
| Muc | = | 778.36 x (0.5 x 450 - 0.333 x 450) = 58.45 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 58.45 + (22.28) |
| = | 80.73 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00285 | 352.60 | 8.92 | 343.68 | 194.35 | 104 | 20.21 |
| 2 | 565 | -0.00008 | -15.67 | 0.00 | -15.67 | -8.86 | -104 | 0.92 |
| Total | 185.49 | | 21.13 |
| xuy | = | 248 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 248/300 = 0.298 |
| Puc | = | 0.298 x 20.00 x 450 x 300 |
| = | 804.94 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 804.94 + (185.49) | = | 990.42 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 248/300 |
| = | 0.345 |
| Muc | = | 804.94 x (0.5 x 300 - 0.345 x 300) = 37.55 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 37.55 + (21.13) |
| = | 58.68 kN-m |
| Pu/Puz | = | 0.631 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.631, | an | = | 1.719 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((21.38/80.73)1.719) + ((19.66/58.68)1.719) |
| = 0.254 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG1 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG2 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 291.16 kN |
| MomentX,(Mx) | = | 103.69 kN-m |
| MomentY,(My) | = | 3.84 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C10 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 6.027 kN.m |
| My_MinEccen | = | 5.823 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 103.693 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 5.823 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 103.69/291.16 = 356 mm |
| Actual eccenY | = | My / P = 3.84/291.16 = 13 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(356,21) = 356 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(13,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(356 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 2 nos. + #16 - 6 nos. (1433 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00261 | 347.97 | 8.92 | 339.05 | 136.34 | 177 | 24.13 |
| 2 | 226 | 0.00044 | 87.54 | 3.48 | 84.06 | 19.01 | 60 | 1.13 |
| 3 | 402 | -0.00174 | -313.29 | 0.00 | -313.29 | -125.98 | -58 | 7.26 |
| 4 | 402 | -0.00395 | -360.90 | 0.00 | -360.90 | -145.13 | -177 | 25.69 |
| Total | -115.75 | | 58.22 |
| xux | = | 189 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 189/450 = 0.151 |
| Puc | = | 0.151 x 20.00 x 300 x 450 |
| = | 408.16 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 408.16 + (-115.75) | = | 292.41 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 189/450 |
| = | 0.175 |
| Muc | = | 408.16 x (0.5 x 450 - 0.175 x 450) = 59.75 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 59.75 + (58.22) |
| = | 117.97 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 716 | 0.00184 | 319.97 | 8.86 | 311.11 | 222.84 | 102 | 22.73 |
| 2 | 716 | -0.00520 | -360.90 | 0.00 | -360.90 | -258.51 | -102 | 26.37 |
| Total | -35.67 | | 49.10 |
| xuy | = | 101 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 101/300 = 0.122 |
| Puc | = | 0.122 x 20.00 x 450 x 300 |
| = | 328.43 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 328.43 + (-35.67) | = | 292.76 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 101/300 |
| = | 0.141 |
| Muc | = | 328.43 x (0.5 x 300 - 0.141 x 300) = 35.41 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 35.41 + (49.10) |
| = | 84.51 kN-m |
| Pu/Puz | = | 0.177 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.177, | an | = | 1.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((103.69/117.97)1.000) + ((5.82/84.51)1.000) |
| = 0.948 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG2 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 2 nos. + #16 - 6 nos. (1433 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 595.50 kN |
| MomentX,(Mx) | = | 93.64 kN-m |
| MomentY,(My) | = | 2.75 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C10 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 12.327 kN.m |
| My_MinEccen | = | 11.910 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 93.636 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 11.910 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 93.64/595.50 = 157 mm |
| Actual eccenY | = | My / P = 2.75/595.50 = 5 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(157,21) = 157 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(5,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(157 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00286 | 352.71 | 8.92 | 343.79 | 77.76 | 179 | 13.92 |
| 2 | 226 | 0.00163 | 306.36 | 8.61 | 297.75 | 67.35 | 90 | 6.03 |
| 3 | 226 | 0.00039 | 77.78 | 3.13 | 74.65 | 16.88 | 0 | 0.00 |
| 4 | 226 | -0.00085 | -169.73 | 0.00 | -169.73 | -38.39 | -90 | 3.44 |
| 5 | 226 | -0.00209 | -330.90 | 0.00 | -330.90 | -74.85 | -179 | 13.40 |
| Total | 48.76 | | 36.78 |
| xux | = | 253 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 253/450 = 0.203 |
| Puc | = | 0.203 x 20.00 x 300 x 450 |
| = | 546.75 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 546.75 + (48.76) | = | 595.51 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 253/450 |
| = | 0.234 |
| Muc | = | 546.75 x (0.5 x 450 - 0.234 x 450) = 65.45 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 65.45 + (36.78) |
| = | 102.23 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00259 | 347.43 | 8.92 | 338.51 | 191.42 | 104 | 19.91 |
| 2 | 565 | -0.00152 | -296.65 | 0.00 | -296.65 | -167.75 | -104 | 17.45 |
| Total | 23.67 | | 37.35 |
| xuy | = | 177 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 177/300 = 0.212 |
| Puc | = | 0.212 x 20.00 x 450 x 300 |
| = | 573.33 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 573.33 + (23.67) | = | 597.00 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 177/300 |
| = | 0.245 |
| Muc | = | 573.33 x (0.5 x 300 - 0.245 x 300) = 43.80 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 43.80 + (37.35) |
| = | 81.15 kN-m |
| Pu/Puz | = | 0.383 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.383, | an | = | 1.304 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((93.64/102.23)1.304) + ((11.91/81.15)1.304) |
| = 0.974 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG2 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 916.58 kN |
| MomentX,(Mx) | = | 92.91 kN-m |
| MomentY,(My) | = | 1.85 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C10 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 18.973 kN.m |
| My_MinEccen | = | 18.332 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 92.914 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 18.332 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 92.91/916.58 = 101 mm |
| Actual eccenY | = | My / P = 1.85/916.58 = 2 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(101,21) = 101 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(2,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(101 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00299 | 353.83 | 8.92 | 344.91 | 138.70 | 177 | 24.55 |
| 2 | 226 | 0.00204 | 329.35 | 8.92 | 320.43 | 72.48 | 88 | 6.34 |
| 3 | 226 | 0.00112 | 223.40 | 7.18 | 216.22 | 48.91 | 0 | 0.00 |
| 4 | 226 | 0.00019 | 38.06 | 1.62 | 36.44 | 8.24 | -88 | -0.72 |
| 5 | 402 | -0.00076 | -151.52 | 0.00 | -151.52 | -60.93 | -177 | 10.78 |
| Total | 207.40 | | 40.95 |
| xux | = | 330 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 330/450 = 0.264 |
| Puc | = | 0.264 x 20.00 x 300 x 450 |
| = | 713.81 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 713.81 + (207.40) | = | 921.21 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 330/450 |
| = | 0.306 |
| Muc | = | 713.81 x (0.5 x 450 - 0.306 x 450) = 62.48 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 62.48 + (40.95) |
| = | 103.43 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 741 | 0.00275 | 351.63 | 8.92 | 342.71 | 254.09 | 102 | 25.92 |
| 2 | 741 | -0.00042 | -84.00 | 0.00 | -84.00 | -62.28 | -102 | 6.35 |
| Total | 191.81 | | 32.27 |
| xuy | = | 225 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 225/300 = 0.270 |
| Puc | = | 0.270 x 20.00 x 450 x 300 |
| = | 729.00 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 729.00 + (191.81) | = | 920.81 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 225/300 |
| = | 0.312 |
| Muc | = | 729.00 x (0.5 x 300 - 0.312 x 300) = 41.12 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 41.12 + (32.27) |
| = | 73.39 kN-m |
| Pu/Puz | = | 0.551 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.551, | an | = | 1.585 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((92.91/103.43)1.585) + ((18.33/73.39)1.585) |
| = 0.955 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG2 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1258.60 kN |
| MomentX,(Mx) | = | 91.36 kN-m |
| MomentY,(My) | = | 0.97 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C10 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 26.053 kN.m |
| My_MinEccen | = | 25.172 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 91.364 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 25.172 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 91.36/1258.60 = 73 mm |
| Actual eccenY | = | My / P = 0.97/1258.60 = 1 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(73,21) = 73 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(73 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 12 nos. (2413 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00304 | 354.25 | 8.92 | 345.33 | 138.87 | 177 | 24.58 |
| 2 | 402 | 0.00236 | 341.07 | 8.92 | 332.14 | 133.56 | 106 | 14.18 |
| 3 | 402 | 0.00169 | 310.13 | 8.70 | 301.44 | 121.21 | 35 | 4.29 |
| 4 | 402 | 0.00101 | 201.46 | 6.72 | 194.73 | 78.31 | -35 | -2.77 |
| 5 | 402 | 0.00033 | 65.91 | 2.70 | 63.21 | 25.42 | -106 | -2.70 |
| 6 | 402 | -0.00035 | -69.64 | 0.00 | -69.64 | -28.00 | -177 | 4.96 |
| Total | 469.37 | | 42.54 |
| xux | = | 366 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 366/450 = 0.293 |
| Puc | = | 0.293 x 20.00 x 300 x 450 |
| = | 789.75 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 789.75 + (469.37) | = | 1259.12 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 366/450 |
| = | 0.338 |
| Muc | = | 789.75 x (0.5 x 450 - 0.338 x 450) = 57.57 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 57.57 + (42.54) |
| = | 100.11 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1206 | 0.00285 | 352.57 | 8.92 | 343.65 | 414.57 | 102 | 42.29 |
| 2 | 1206 | 0.00008 | 15.78 | 0.69 | 15.09 | 18.21 | -102 | -1.86 |
| Total | 432.78 | | 40.43 |
| xuy | = | 258 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 258/300 = 0.309 |
| Puc | = | 0.309 x 20.00 x 450 x 300 |
| = | 835.31 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 835.31 + (432.78) | = | 1268.09 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 258/300 |
| = | 0.358 |
| Muc | = | 835.31 x (0.5 x 300 - 0.358 x 300) = 35.71 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 35.71 + (40.43) |
| = | 76.14 kN-m |
| Pu/Puz | = | 0.647 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.647, | an | = | 1.746 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((91.36/100.11)1.746) + ((25.17/76.14)1.746) |
| = 0.997 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG2 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 12 nos. (2413 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3375 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1639.13 kN |
| MomentX,(Mx) | = | 66.74 kN-m |
| MomentY,(My) | = | 0.99 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C10 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 35.651 kN.m |
| My_MinEccen | = | 32.783 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 66.735 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 32.783 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 66.74/1639.13 = 41 mm |
| Actual eccenY | = | My / P = 0.99/1639.13 = 1 mm |
| eccenXMin | = | (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(41,22) = 41 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(41 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00309 | 354.70 | 8.92 | 345.78 | 139.05 | 177 | 24.61 |
| 2 | 402 | 0.00266 | 349.26 | 8.92 | 340.34 | 136.86 | 126 | 17.30 |
| 3 | 402 | 0.00223 | 336.22 | 8.92 | 327.30 | 131.62 | 76 | 9.98 |
| 4 | 402 | 0.00180 | 317.35 | 8.83 | 308.52 | 124.06 | 25 | 3.14 |
| 5 | 402 | 0.00137 | 274.06 | 8.04 | 266.03 | 106.98 | -25 | -2.70 |
| 6 | 402 | 0.00094 | 188.00 | 6.41 | 181.59 | 73.02 | -76 | -5.54 |
| 7 | 402 | 0.00051 | 101.94 | 3.97 | 97.97 | 39.40 | -126 | -4.98 |
| 8 | 402 | 0.00008 | 15.87 | 0.69 | 15.18 | 6.10 | -177 | -1.08 |
| Total | 757.08 | | 40.73 |
| xux | = | 411 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 411/450 = 0.329 |
| Puc | = | 0.329 x 20.00 x 300 x 450 |
| = | 888.47 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 888.47 + (757.08) | = | 1645.55 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 411/450 |
| = | 0.380 |
| Muc | = | 888.47 x (0.5 x 450 - 0.380 x 450) = 47.88 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 47.88 + (40.73) |
| = | 88.61 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1608 | 0.00292 | 353.22 | 8.92 | 344.30 | 553.80 | 102 | 56.49 |
| 2 | 1608 | 0.00047 | 93.03 | 3.67 | 89.37 | 143.74 | -102 | -14.66 |
| Total | 697.54 | | 41.83 |
| xuy | = | 291 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 291/300 = 0.349 |
| Puc | = | 0.349 x 20.00 x 450 x 300 |
| = | 941.63 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 941.63 + (697.54) | = | 1639.17 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 291/300 |
| = | 0.403 |
| Muc | = | 941.63 x (0.5 x 300 - 0.403 x 300) = 27.40 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 27.40 + (41.83) |
| = | 69.23 kN-m |
| Pu/Puz | = | 0.749 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.749, | an | = | 1.916 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((66.74/88.61)1.916) + ((32.78/69.23)1.916) |
| = 0.820 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG2 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG3 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 137.99 kN |
| MomentX,(Mx) | = | 49.09 kN-m |
| MomentY,(My) | = | 26.54 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C11 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 2.856 kN.m |
| My_MinEccen | = | 2.760 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 49.086 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 26.543 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 49.09/137.99 = 356 mm |
| Actual eccenY | = | My / P = 26.54/137.99 = 192 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(356,21) = 356 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(192,20) = 192 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(356 mm) > 0.05 x 300(15 mm)
and eccenY(192 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 6 nos. (1206 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00225 | 336.80 | 8.92 | 327.88 | 131.85 | 177 | 23.34 |
| 2 | 402 | -0.00238 | -341.53 | 0.00 | -341.53 | -137.34 | 0 | 0.00 |
| 3 | 402 | -0.00700 | -360.90 | 0.00 | -360.90 | -145.13 | -177 | 25.69 |
| Total | -150.62 | | 49.02 |
| xux | = | 134 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 134/450 = 0.107 |
| Puc | = | 0.107 x 20.00 x 300 x 450 |
| = | 289.51 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 289.51 + (-150.62) | = | 138.90 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 134/450 |
| = | 0.124 |
| Muc | = | 289.51 x (0.5 x 450 - 0.124 x 450) = 49.00 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 49.00 + (49.02) |
| = | 98.02 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 603 | 0.00111 | 221.14 | 7.14 | 214.00 | 129.08 | 102 | 13.17 |
| 2 | 603 | -0.00907 | -360.90 | 0.00 | -360.90 | -217.69 | -102 | 22.20 |
| Total | -88.61 | | 35.37 |
| xuy | = | 70 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 70/300 = 0.084 |
| Puc | = | 0.084 x 20.00 x 450 x 300 |
| = | 227.34 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 227.34 + (-88.61) | = | 138.73 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 70/300 |
| = | 0.097 |
| Muc | = | 227.34 x (0.5 x 300 - 0.097 x 300) = 27.46 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 27.46 + (35.37) |
| = | 62.84 kN-m |
| Pu/Puz | = | 0.087 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.087, | an | = | 1.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((49.09/98.02)1.000) + ((26.54/62.84)1.000) |
| = 0.923 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG3 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 6 nos. (1206 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 323.16 kN |
| MomentX,(Mx) | = | 45.84 kN-m |
| MomentY,(My) | = | 23.18 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C11 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 6.689 kN.m |
| My_MinEccen | = | 6.463 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 45.835 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 23.180 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 45.84/323.16 = 142 mm |
| Actual eccenY | = | My / P = 23.18/323.16 = 72 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(142,21) = 142 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(72,20) = 72 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(142 mm) > 0.05 x 300(15 mm)
and eccenY(72 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00263 | 348.50 | 8.92 | 339.58 | 76.81 | 179 | 13.75 |
| 2 | 226 | 0.00094 | 188.54 | 6.43 | 182.11 | 41.19 | 90 | 3.69 |
| 3 | 226 | -0.00075 | -149.29 | 0.00 | -149.29 | -33.77 | 0 | 0.00 |
| 4 | 226 | -0.00244 | -343.46 | 0.00 | -343.46 | -77.69 | -90 | 6.95 |
| 5 | 226 | -0.00412 | -360.90 | 0.00 | -360.90 | -81.63 | -179 | 14.61 |
| Total | -75.09 | | 39.00 |
| xux | = | 185 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 185/450 = 0.148 |
| Puc | = | 0.148 x 20.00 x 300 x 450 |
| = | 400.57 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 400.57 + (-75.09) | = | 325.48 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 185/450 |
| = | 0.171 |
| Muc | = | 400.57 x (0.5 x 450 - 0.171 x 450) = 59.23 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 59.23 + (39.00) |
| = | 98.23 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00201 | 327.98 | 8.92 | 319.06 | 180.43 | 104 | 18.76 |
| 2 | 565 | -0.00475 | -360.90 | 0.00 | -360.90 | -204.08 | -104 | 21.22 |
| Total | -23.66 | | 39.99 |
| xuy | = | 108 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 108/300 = 0.129 |
| Puc | = | 0.129 x 20.00 x 450 x 300 |
| = | 349.31 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 349.31 + (-23.66) | = | 325.65 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 108/300 |
| = | 0.149 |
| Muc | = | 349.31 x (0.5 x 300 - 0.149 x 300) = 36.73 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 36.73 + (39.99) |
| = | 76.72 kN-m |
| Pu/Puz | = | 0.208 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.208, | an | = | 1.013 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((45.84/98.23)1.013) + ((23.18/76.72)1.013) |
| = 0.760 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG3 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 491.14 kN |
| MomentX,(Mx) | = | 38.31 kN-m |
| MomentY,(My) | = | 19.18 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C11 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 10.167 kN.m |
| My_MinEccen | = | 9.823 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 38.309 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 19.176 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 38.31/491.14 = 78 mm |
| Actual eccenY | = | My / P = 19.18/491.14 = 39 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(78,21) = 78 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(39,20) = 39 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(78 mm) > 0.05 x 300(15 mm)
and eccenY(39 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00279 | 352.09 | 8.92 | 343.17 | 77.62 | 179 | 13.89 |
| 2 | 226 | 0.00142 | 283.33 | 8.16 | 275.17 | 62.24 | 90 | 5.57 |
| 3 | 226 | 0.00004 | 8.11 | 0.36 | 7.75 | 1.75 | 0 | 0.00 |
| 4 | 226 | -0.00134 | -267.11 | 0.00 | -267.11 | -60.42 | -90 | 5.41 |
| 5 | 226 | -0.00271 | -350.56 | 0.00 | -350.56 | -79.29 | -179 | 14.19 |
| Total | 1.90 | | 39.07 |
| xux | = | 228 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 228/450 = 0.182 |
| Puc | = | 0.182 x 20.00 x 300 x 450 |
| = | 491.70 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 491.70 + (1.90) | = | 493.60 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 228/450 |
| = | 0.210 |
| Muc | = | 491.70 x (0.5 x 450 - 0.210 x 450) = 64.07 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 64.07 + (39.07) |
| = | 103.14 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00245 | 343.86 | 8.92 | 334.94 | 189.40 | 104 | 19.70 |
| 2 | 565 | -0.00229 | -338.43 | 0.00 | -338.43 | -191.38 | -104 | 19.90 |
| Total | -1.97 | | 39.60 |
| xuy | = | 154 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 154/300 = 0.184 |
| Puc | = | 0.184 x 20.00 x 450 x 300 |
| = | 497.39 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 497.39 + (-1.97) | = | 495.42 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 154/300 |
| = | 0.213 |
| Muc | = | 497.39 x (0.5 x 300 - 0.213 x 300) = 42.84 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 42.84 + (39.60) |
| = | 82.45 kN-m |
| Pu/Puz | = | 0.315 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.315, | an | = | 1.192 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((38.31/103.14)1.192) + ((19.18/82.45)1.192) |
| = 0.483 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG3 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 740.97 kN |
| MomentX,(Mx) | = | 27.01 kN-m |
| MomentY,(My) | = | 13.81 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C11 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 15.338 kN.m |
| My_MinEccen | = | 14.819 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 27.009 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 14.819 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 27.01/740.97 = 36 mm |
| Actual eccenY | = | My / P = 13.81/740.97 = 19 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(36,21) = 36 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(19,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(36 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00295 | 353.48 | 8.92 | 344.56 | 77.94 | 179 | 13.95 |
| 2 | 226 | 0.00188 | 322.58 | 8.89 | 313.69 | 70.96 | 90 | 6.35 |
| 3 | 226 | 0.00082 | 163.47 | 5.80 | 157.67 | 35.66 | 0 | 0.00 |
| 4 | 226 | -0.00025 | -49.95 | 0.00 | -49.95 | -11.30 | -90 | 1.01 |
| 5 | 226 | -0.00132 | -263.36 | 0.00 | -263.36 | -59.57 | -179 | 10.66 |
| Total | 113.69 | | 31.98 |
| xux | = | 294 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 294/450 = 0.235 |
| Puc | = | 0.235 x 20.00 x 300 x 450 |
| = | 634.08 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 634.08 + (113.69) | = | 747.77 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 294/450 |
| = | 0.271 |
| Muc | = | 634.08 x (0.5 x 450 - 0.271 x 450) = 65.23 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 65.23 + (31.98) |
| = | 97.21 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00270 | 350.29 | 8.92 | 341.37 | 193.04 | 104 | 20.08 |
| 2 | 565 | -0.00091 | -182.11 | 0.00 | -182.11 | -102.98 | -104 | 10.71 |
| Total | 90.06 | | 30.79 |
| xuy | = | 202 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 202/300 = 0.242 |
| Puc | = | 0.242 x 20.00 x 450 x 300 |
| = | 653.06 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 653.06 + (90.06) | = | 743.12 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 202/300 |
| = | 0.280 |
| Muc | = | 653.06 x (0.5 x 300 - 0.280 x 300) = 43.20 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 43.20 + (30.79) |
| = | 73.99 kN-m |
| Pu/Puz | = | 0.476 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.476, | an | = | 1.460 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((27.01/97.21)1.460) + ((14.82/73.99)1.460) |
| = 0.250 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG3 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3375 mm |
From analysis results,loads on column
| Axial load,(P) | = | 854.17 kN |
| MomentX,(Mx) | = | 15.00 kN-m |
| MomentY,(My) | = | 6.47 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C11 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 18.578 kN.m |
| My_MinEccen | = | 17.083 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 18.578 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 17.083 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 15.00/854.17 = 18 mm |
| Actual eccenY | = | My / P = 6.47/854.17 = 8 mm |
| eccenXMin | = | (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(18,22) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(8,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00300 | 353.92 | 8.92 | 345.00 | 78.04 | 179 | 13.97 |
| 2 | 226 | 0.00203 | 328.98 | 8.92 | 320.06 | 72.40 | 90 | 6.48 |
| 3 | 226 | 0.00107 | 213.04 | 6.97 | 206.07 | 46.61 | 0 | 0.00 |
| 4 | 226 | 0.00010 | 19.34 | 0.84 | 18.50 | 4.18 | -90 | -0.37 |
| 5 | 226 | -0.00087 | -174.36 | 0.00 | -174.36 | -39.44 | -179 | 7.06 |
| Total | 161.79 | | 27.13 |
| xux | = | 323 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 323/450 = 0.259 |
| Puc | = | 0.259 x 20.00 x 300 x 450 |
| = | 698.63 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 698.63 + (161.79) | = | 860.42 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 323/450 |
| = | 0.299 |
| Muc | = | 698.63 x (0.5 x 450 - 0.299 x 450) = 63.19 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 63.19 + (27.13) |
| = | 90.32 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00278 | 351.95 | 8.92 | 343.03 | 193.98 | 104 | 20.17 |
| 2 | 565 | -0.00049 | -98.54 | 0.00 | -98.54 | -55.72 | -104 | 5.80 |
| Total | 138.25 | | 25.97 |
| xuy | = | 223 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 223/300 = 0.267 |
| Puc | = | 0.267 x 20.00 x 450 x 300 |
| = | 721.41 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 721.41 + (138.25) | = | 859.66 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 223/300 |
| = | 0.309 |
| Muc | = | 721.41 x (0.5 x 300 - 0.309 x 300) = 41.39 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 41.39 + (25.97) |
| = | 67.36 kN-m |
| Pu/Puz | = | 0.549 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.549, | an | = | 1.581 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((18.58/90.32)1.581) + ((17.08/67.36)1.581) |
| = 0.196 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG3 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG4 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3450 mm |
From analysis results,loads on column
| Axial load,(P) | = | 286.83 kN |
| MomentX,(Mx) | = | 62.42 kN-m |
| MomentY,(My) | = | 63.98 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C12 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 6.282 kN.m |
| My_MinEccen | = | 5.737 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 62.421 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 63.978 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 62.42/286.83 = 218 mm |
| Actual eccenY | = | My / P = 63.98/286.83 = 223 mm |
| eccenXMin | = | (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(218,22) = 218 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(223,20) = 223 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(218 mm) > 0.05 x 300(15 mm)
and eccenY(223 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 12 nos. (2413 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00264 | 348.59 | 8.92 | 339.67 | 136.59 | 177 | 24.18 |
| 2 | 402 | 0.00136 | 271.87 | 8.00 | 263.86 | 106.10 | 106 | 11.27 |
| 3 | 402 | 0.00008 | 16.72 | 0.73 | 15.99 | 6.43 | 35 | 0.23 |
| 4 | 402 | -0.00119 | -238.43 | 0.00 | -238.43 | -95.88 | -35 | 3.39 |
| 5 | 402 | -0.00247 | -344.29 | 0.00 | -344.29 | -138.45 | -106 | 14.70 |
| 6 | 402 | -0.00374 | -360.41 | 0.00 | -360.41 | -144.93 | -177 | 25.65 |
| Total | -130.13 | | 79.42 |
| xux | = | 194 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 194/450 = 0.155 |
| Puc | = | 0.155 x 20.00 x 300 x 450 |
| = | 419.55 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 419.55 + (-130.13) | = | 289.42 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 194/450 |
| = | 0.180 |
| Muc | = | 419.55 x (0.5 x 450 - 0.180 x 450) = 60.50 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 60.50 + (79.42) |
| = | 139.92 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1206 | 0.00192 | 324.54 | 8.90 | 315.64 | 380.78 | 102 | 38.84 |
| 2 | 1206 | -0.00482 | -360.90 | 0.00 | -360.90 | -435.38 | -102 | 44.41 |
| Total | -54.60 | | 83.25 |
| xuy | = | 106 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 106/300 = 0.127 |
| Puc | = | 0.127 x 20.00 x 450 x 300 |
| = | 343.62 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 343.62 + (-54.60) | = | 289.02 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 106/300 |
| = | 0.147 |
| Muc | = | 343.62 x (0.5 x 300 - 0.147 x 300) = 36.38 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 36.38 + (83.25) |
| = | 119.63 kN-m |
| Pu/Puz | = | 0.148 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.148, | an | = | 1.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((62.42/139.92)1.000) + ((63.98/119.63)1.000) |
| = 0.981 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG4 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 12 nos. (2413 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3450 mm |
From analysis results,loads on column
| Axial load,(P) | = | 594.33 kN |
| MomentX,(Mx) | = | 42.37 kN-m |
| MomentY,(My) | = | 56.23 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C12 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 13.016 kN.m |
| My_MinEccen | = | 11.887 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 42.366 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 56.229 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 42.37/594.33 = 71 mm |
| Actual eccenY | = | My / P = 56.23/594.33 = 95 mm |
| eccenXMin | = | (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(71,22) = 71 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(95,20) = 95 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(71 mm) > 0.05 x 300(15 mm)
and eccenY(95 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00286 | 352.71 | 8.92 | 343.79 | 77.76 | 179 | 13.92 |
| 2 | 226 | 0.00163 | 306.36 | 8.61 | 297.75 | 67.35 | 90 | 6.03 |
| 3 | 226 | 0.00039 | 77.78 | 3.13 | 74.65 | 16.88 | 0 | 0.00 |
| 4 | 226 | -0.00085 | -169.73 | 0.00 | -169.73 | -38.39 | -90 | 3.44 |
| 5 | 226 | -0.00209 | -330.90 | 0.00 | -330.90 | -74.85 | -179 | 13.40 |
| Total | 48.76 | | 36.78 |
| xux | = | 253 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 253/450 = 0.203 |
| Puc | = | 0.203 x 20.00 x 300 x 450 |
| = | 546.75 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 546.75 + (48.76) | = | 595.51 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 253/450 |
| = | 0.234 |
| Muc | = | 546.75 x (0.5 x 450 - 0.234 x 450) = 65.45 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 65.45 + (36.78) |
| = | 102.23 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00259 | 347.43 | 8.92 | 338.51 | 191.42 | 104 | 19.91 |
| 2 | 565 | -0.00152 | -296.65 | 0.00 | -296.65 | -167.75 | -104 | 17.45 |
| Total | 23.67 | | 37.35 |
| xuy | = | 177 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 177/300 = 0.212 |
| Puc | = | 0.212 x 20.00 x 450 x 300 |
| = | 573.33 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 573.33 + (23.67) | = | 597.00 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 177/300 |
| = | 0.245 |
| Muc | = | 573.33 x (0.5 x 300 - 0.245 x 300) = 43.80 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 43.80 + (37.35) |
| = | 81.15 kN-m |
| Pu/Puz | = | 0.382 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.382, | an | = | 1.303 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((42.37/102.23)1.303) + ((56.23/81.15)1.303) |
| = 0.937 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG4 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3450 mm |
From analysis results,loads on column
| Axial load,(P) | = | 894.06 kN |
| MomentX,(Mx) | = | 55.58 kN-m |
| MomentY,(My) | = | 55.13 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C12 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 19.580 kN.m |
| My_MinEccen | = | 17.881 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 55.578 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 55.133 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 55.58/894.06 = 62 mm |
| Actual eccenY | = | My / P = 55.13/894.06 = 62 mm |
| eccenXMin | = | (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(62,22) = 62 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(62,20) = 62 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(62 mm) > 0.05 x 300(15 mm)
and eccenY(62 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00298 | 353.75 | 8.92 | 344.83 | 138.67 | 177 | 24.54 |
| 2 | 226 | 0.00202 | 328.48 | 8.92 | 319.56 | 72.28 | 88 | 6.32 |
| 3 | 226 | 0.00108 | 215.68 | 7.03 | 208.65 | 47.20 | 0 | 0.00 |
| 4 | 226 | 0.00014 | 27.33 | 1.18 | 26.15 | 5.92 | -88 | -0.52 |
| 5 | 402 | -0.00083 | -165.33 | 0.00 | -165.33 | -66.48 | -177 | 11.77 |
| Total | 197.58 | | 42.12 |
| xux | = | 325 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 325/450 = 0.260 |
| Puc | = | 0.260 x 20.00 x 300 x 450 |
| = | 702.42 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 702.42 + (197.58) | = | 900.00 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 325/450 |
| = | 0.301 |
| Muc | = | 702.42 x (0.5 x 450 - 0.301 x 450) = 63.02 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 63.02 + (42.12) |
| = | 105.14 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 741 | 0.00274 | 351.32 | 8.92 | 342.40 | 253.86 | 102 | 25.89 |
| 2 | 741 | -0.00048 | -96.44 | 0.00 | -96.44 | -71.51 | -102 | 7.29 |
| Total | 182.36 | | 33.19 |
| xuy | = | 221 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 221/300 = 0.266 |
| Puc | = | 0.266 x 20.00 x 450 x 300 |
| = | 717.61 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 717.61 + (182.36) | = | 899.97 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 221/300 |
| = | 0.307 |
| Muc | = | 717.61 x (0.5 x 300 - 0.307 x 300) = 41.52 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 41.52 + (33.19) |
| = | 74.71 kN-m |
| Pu/Puz | = | 0.538 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.538, | an | = | 1.563 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((55.58/105.14)1.563) + ((55.13/74.71)1.563) |
| = 0.991 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG4 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3450 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1289.95 kN |
| MomentX,(Mx) | = | 51.02 kN-m |
| MomentY,(My) | = | 56.42 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C12 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 28.250 kN.m |
| My_MinEccen | = | 25.799 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 51.019 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 56.421 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 51.02/1289.95 = 40 mm |
| Actual eccenY | = | My / P = 56.42/1289.95 = 44 mm |
| eccenXMin | = | (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(40,22) = 40 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(44,20) = 44 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(40 mm) > 0.05 x 300(15 mm)
and eccenY(44 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 12 nos. (2413 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00305 | 354.35 | 8.92 | 345.43 | 138.91 | 177 | 24.59 |
| 2 | 402 | 0.00239 | 342.05 | 8.92 | 333.13 | 133.96 | 106 | 14.23 |
| 3 | 402 | 0.00173 | 312.79 | 8.75 | 304.04 | 122.26 | 35 | 4.33 |
| 4 | 402 | 0.00107 | 213.16 | 6.97 | 206.19 | 82.91 | -35 | -2.94 |
| 5 | 402 | 0.00040 | 80.79 | 3.24 | 77.55 | 31.19 | -106 | -3.31 |
| 6 | 402 | -0.00026 | -51.57 | 0.00 | -51.57 | -20.74 | -177 | 3.67 |
| Total | 488.48 | | 40.56 |
| xux | = | 374 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 374/450 = 0.300 |
| Puc | = | 0.300 x 20.00 x 300 x 450 |
| = | 808.73 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 808.73 + (488.48) | = | 1297.22 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 374/450 |
| = | 0.346 |
| Muc | = | 808.73 x (0.5 x 450 - 0.346 x 450) = 56.00 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 56.00 + (40.56) |
| = | 96.56 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1206 | 0.00286 | 352.67 | 8.92 | 343.75 | 414.70 | 102 | 42.30 |
| 2 | 1206 | 0.00014 | 28.00 | 1.21 | 26.79 | 32.32 | -102 | -3.30 |
| Total | 447.02 | | 39.00 |
| xuy | = | 263 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 263/300 = 0.315 |
| Puc | = | 0.315 x 20.00 x 450 x 300 |
| = | 850.50 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 850.50 + (447.02) | = | 1297.52 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 263/300 |
| = | 0.364 |
| Muc | = | 850.50 x (0.5 x 300 - 0.364 x 300) = 34.70 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 34.70 + (39.00) |
| = | 73.70 kN-m |
| Pu/Puz | = | 0.663 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.663, | an | = | 1.772 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((51.02/96.56)1.772) + ((56.42/73.70)1.772) |
| = 0.946 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG4 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 12 nos. (2413 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3675 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1564.72 kN |
| MomentX,(Mx) | = | 19.13 kN-m |
| MomentY,(My) | = | 39.26 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C12 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3675 x 1.000 ) / 450 = 8.17 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3675 x 1.000 ) / 300 = 12.25 > 12.00, |
column is slender in this direction.| Puz | = | (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast |
| = | 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 2814.87 = 2065793.701 N |
| k1 | = | 0.193600 |
| k2 | = | 0.032400 MPa |
| Pub | = | (k1 x fck x Ag) + (k2 x Ast x 100) |
| = | (0.193600 x 20.00 x 135000.01) + (0.03 x 2814.87 x 100) = 531840.210 N |
| k | = | (Puz - Pu) / (Puz - Pub) <= 1 |
| = | (2065793.701 - 1564720.581) / (2065793.701 - 531840.210) = 0.326655 |
| MuaddY | = | Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k |
| = | 1564720.58 x (300.00/2000) x (3675.00/300.00) x (3675.00/300.00) x 0.326655 |
| = | 11505067.825 N.mm = 11.505 kN.m |
| Mx_MinEccen | = | 34.972 kN.m |
| My_MinEccen | = | 31.294 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 34.972 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 50.764 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 19.13/1564.72 = 12 mm |
| Actual eccenY | = | My / P = 39.26/1564.72 = 25 mm |
| eccenXMin | = | (L/500) + (D/30) = (3675/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3675/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(12,22) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(25,20) = 25 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00310 | 354.76 | 8.92 | 345.84 | 139.07 | 177 | 24.62 |
| 2 | 402 | 0.00260 | 347.81 | 8.92 | 338.89 | 136.28 | 118 | 16.08 |
| 3 | 402 | 0.00211 | 331.83 | 8.92 | 322.91 | 129.85 | 59 | 7.66 |
| 4 | 402 | 0.00162 | 305.53 | 8.59 | 296.94 | 119.40 | 0 | 0.00 |
| 5 | 402 | 0.00112 | 224.81 | 7.21 | 217.60 | 87.50 | -59 | -5.16 |
| 6 | 402 | 0.00063 | 126.09 | 4.74 | 121.35 | 48.80 | -118 | -5.76 |
| 7 | 402 | 0.00014 | 27.37 | 1.18 | 26.19 | 10.53 | -177 | -1.86 |
| Total | 671.43 | | 35.57 |
| xux | = | 418 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 418/450 = 0.335 |
| Puc | = | 0.335 x 20.00 x 300 x 450 |
| = | 903.66 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 903.66 + (671.43) | = | 1575.09 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 418/450 |
| = | 0.387 |
| Muc | = | 903.66 x (0.5 x 450 - 0.387 x 450) = 46.05 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 46.05 + (35.57) |
| = | 81.62 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1407 | 0.00293 | 353.26 | 8.92 | 344.34 | 484.63 | 102 | 49.43 |
| 2 | 1407 | 0.00049 | 97.89 | 3.83 | 94.06 | 132.38 | -102 | -13.50 |
| Total | 617.01 | | 35.93 |
| xuy | = | 293 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 293/300 = 0.352 |
| Puc | = | 0.352 x 20.00 x 450 x 300 |
| = | 949.22 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 949.22 + (617.01) | = | 1566.23 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 293/300 |
| = | 0.406 |
| Muc | = | 949.22 x (0.5 x 300 - 0.406 x 300) = 26.70 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 26.70 + (35.93) |
| = | 62.63 kN-m |
| Pu/Puz | = | 0.757 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.757, | an | = | 1.929 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((34.97/81.62)1.929) + ((50.76/62.63)1.929) |
| = 0.862 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG4 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG5 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 296.80 kN |
| MomentX,(Mx) | = | 41.32 kN-m |
| MomentY,(My) | = | 64.49 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C13 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 6.144 kN.m |
| My_MinEccen | = | 5.936 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 41.317 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 64.493 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 41.32/296.80 = 139 mm |
| Actual eccenY | = | My / P = 64.49/296.80 = 217 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(139,21) = 139 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(217,20) = 217 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(139 mm) > 0.05 x 300(15 mm)
and eccenY(217 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. (1810 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00266 | 349.12 | 8.92 | 340.20 | 76.95 | 179 | 13.77 |
| 2 | 226 | 0.00172 | 312.15 | 8.74 | 303.41 | 68.63 | 128 | 8.77 |
| 3 | 226 | 0.00078 | 155.75 | 5.59 | 150.16 | 33.97 | 77 | 2.61 |
| 4 | 226 | -0.00016 | -31.95 | 0.00 | -31.95 | -7.23 | 26 | -0.18 |
| 5 | 226 | -0.00110 | -219.66 | 0.00 | -219.66 | -49.69 | -26 | 1.27 |
| 6 | 226 | -0.00204 | -329.09 | 0.00 | -329.09 | -74.44 | -77 | 5.71 |
| 7 | 226 | -0.00298 | -353.68 | 0.00 | -353.68 | -80.00 | -128 | 10.23 |
| 8 | 226 | -0.00391 | -360.90 | 0.00 | -360.90 | -81.63 | -179 | 14.61 |
| Total | -113.44 | | 56.79 |
| xux | = | 191 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 191/450 = 0.153 |
| Puc | = | 0.153 x 20.00 x 300 x 450 |
| = | 411.96 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 411.96 + (-113.44) | = | 298.52 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 191/450 |
| = | 0.176 |
| Muc | = | 411.96 x (0.5 x 450 - 0.176 x 450) = 60.01 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 60.01 + (56.79) |
| = | 116.80 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 905 | 0.00196 | 326.13 | 8.92 | 317.22 | 287.01 | 104 | 29.85 |
| 2 | 905 | -0.00502 | -360.90 | 0.00 | -360.90 | -326.53 | -104 | 33.96 |
| Total | -39.52 | | 63.81 |
| xuy | = | 104 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 104/300 = 0.125 |
| Puc | = | 0.125 x 20.00 x 450 x 300 |
| = | 337.92 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 337.92 + (-39.52) | = | 298.40 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 104/300 |
| = | 0.145 |
| Muc | = | 337.92 x (0.5 x 300 - 0.145 x 300) = 36.03 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 36.03 + (63.81) |
| = | 99.84 kN-m |
| Pu/Puz | = | 0.168 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.168, | an | = | 1.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((41.32/116.80)1.000) + ((64.49/99.84)1.000) |
| = 1.000 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG5 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. (1810 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 611.13 kN |
| MomentX,(Mx) | = | 35.92 kN-m |
| MomentY,(My) | = | 57.31 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C13 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 12.650 kN.m |
| My_MinEccen | = | 12.223 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 35.918 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 57.310 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 35.92/611.13 = 59 mm |
| Actual eccenY | = | My / P = 57.31/611.13 = 94 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(59,21) = 59 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(94,20) = 94 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(59 mm) > 0.05 x 300(15 mm)
and eccenY(94 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00288 | 352.82 | 8.92 | 343.90 | 77.79 | 179 | 13.92 |
| 2 | 226 | 0.00166 | 308.86 | 8.67 | 300.19 | 67.90 | 90 | 6.08 |
| 3 | 226 | 0.00045 | 90.48 | 3.58 | 86.90 | 19.66 | 0 | 0.00 |
| 4 | 226 | -0.00076 | -151.98 | 0.00 | -151.98 | -34.38 | -90 | 3.08 |
| 5 | 226 | -0.00197 | -326.72 | 0.00 | -326.72 | -73.90 | -179 | 13.23 |
| Total | 57.07 | | 36.31 |
| xux | = | 258 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 258/450 = 0.207 |
| Puc | = | 0.207 x 20.00 x 300 x 450 |
| = | 558.14 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 558.14 + (57.07) | = | 615.21 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 258/450 |
| = | 0.239 |
| Muc | = | 558.14 x (0.5 x 450 - 0.239 x 450) = 65.59 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 65.59 + (36.31) |
| = | 101.89 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00261 | 347.89 | 8.92 | 338.97 | 191.68 | 104 | 19.93 |
| 2 | 565 | -0.00143 | -285.21 | 0.00 | -285.21 | -161.28 | -104 | 16.77 |
| Total | 30.40 | | 36.71 |
| xuy | = | 180 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 180/300 = 0.217 |
| Puc | = | 0.217 x 20.00 x 450 x 300 |
| = | 584.72 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 584.72 + (30.40) | = | 615.12 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 180/300 |
| = | 0.250 |
| Muc | = | 584.72 x (0.5 x 300 - 0.250 x 300) = 43.81 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 43.81 + (36.71) |
| = | 80.52 kN-m |
| Pu/Puz | = | 0.393 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.393, | an | = | 1.321 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((35.92/101.89)1.321) + ((57.31/80.52)1.321) |
| = 0.890 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG5 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 919.74 kN |
| MomentX,(Mx) | = | 32.03 kN-m |
| MomentY,(My) | = | 56.33 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C13 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 19.039 kN.m |
| My_MinEccen | = | 18.395 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 32.026 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 56.329 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 32.03/919.74 = 35 mm |
| Actual eccenY | = | My / P = 56.33/919.74 = 61 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(35,21) = 35 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(61,20) = 61 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(35 mm) > 0.05 x 300(15 mm)
and eccenY(61 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 6 nos. (1206 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00301 | 353.99 | 8.92 | 345.07 | 138.76 | 177 | 24.56 |
| 2 | 402 | 0.00120 | 240.51 | 7.50 | 233.01 | 93.70 | 0 | 0.00 |
| 3 | 402 | -0.00060 | -120.95 | 0.00 | -120.95 | -48.64 | -177 | 8.61 |
| Total | 183.82 | | 33.17 |
| xux | = | 343 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 343/450 = 0.274 |
| Puc | = | 0.274 x 20.00 x 300 x 450 |
| = | 740.39 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 740.39 + (183.82) | = | 924.21 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 343/450 |
| = | 0.317 |
| Muc | = | 740.39 x (0.5 x 450 - 0.317 x 450) = 61.01 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 61.01 + (33.17) |
| = | 94.18 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 603 | 0.00278 | 351.94 | 8.92 | 343.02 | 206.90 | 102 | 21.10 |
| 2 | 603 | -0.00030 | -60.24 | 0.00 | -60.24 | -36.34 | -102 | 3.71 |
| Total | 170.57 | | 24.81 |
| xuy | = | 232 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 232/300 = 0.278 |
| Puc | = | 0.278 x 20.00 x 450 x 300 |
| = | 751.78 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 751.78 + (170.57) | = | 922.35 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 232/300 |
| = | 0.322 |
| Muc | = | 751.78 x (0.5 x 300 - 0.322 x 300) = 40.20 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 40.20 + (24.81) |
| = | 65.01 kN-m |
| Pu/Puz | = | 0.582 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.582, | an | = | 1.637 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((32.03/94.18)1.637) + ((56.33/65.01)1.637) |
| = 0.962 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG5 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 6 nos. (1206 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1216.66 kN |
| MomentX,(Mx) | = | 26.15 kN-m |
| MomentY,(My) | = | 56.89 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C13 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 25.185 kN.m |
| My_MinEccen | = | 24.333 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 26.152 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 56.888 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 26.15/1216.66 = 21 mm |
| Actual eccenY | = | My / P = 56.89/1216.66 = 47 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(21,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(47,20) = 47 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(47 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 6 nos. (1885 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00307 | 354.47 | 8.92 | 345.55 | 138.96 | 177 | 24.60 |
| 2 | 226 | 0.00242 | 343.09 | 8.92 | 334.17 | 75.59 | 106 | 8.00 |
| 3 | 226 | 0.00179 | 316.99 | 8.83 | 308.16 | 69.71 | 37 | 2.55 |
| 4 | 402 | 0.00115 | 230.10 | 7.31 | 222.79 | 89.59 | -35 | -3.10 |
| 5 | 226 | 0.00051 | 101.22 | 3.94 | 97.28 | 22.00 | -106 | -2.33 |
| 6 | 402 | -0.00014 | -27.66 | 0.00 | -27.66 | -11.12 | -177 | 1.97 |
| Total | 384.72 | | 31.68 |
| xux | = | 387 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 387/450 = 0.309 |
| Puc | = | 0.309 x 20.00 x 300 x 450 |
| = | 835.31 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 835.31 + (384.72) | = | 1220.03 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 387/450 |
| = | 0.358 |
| Muc | = | 835.31 x (0.5 x 450 - 0.358 x 450) = 53.56 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 53.56 + (31.68) |
| = | 85.25 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 942 | 0.00287 | 352.77 | 8.92 | 343.85 | 324.07 | 102 | 33.06 |
| 2 | 942 | 0.00020 | 39.79 | 1.69 | 38.10 | 35.91 | -102 | -3.66 |
| Total | 359.99 | | 29.39 |
| xuy | = | 267 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 267/300 = 0.321 |
| Puc | = | 0.321 x 20.00 x 450 x 300 |
| = | 865.69 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 865.69 + (359.99) | = | 1225.67 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 267/300 |
| = | 0.370 |
| Muc | = | 865.69 x (0.5 x 300 - 0.370 x 300) = 33.63 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 33.63 + (29.39) |
| = | 63.02 kN-m |
| Pu/Puz | = | 0.682 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.682, | an | = | 1.803 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((26.15/85.25)1.803) + ((56.89/63.02)1.803) |
| = 0.950 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG5 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 6 nos. (1885 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3375 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1503.07 kN |
| MomentX,(Mx) | = | 13.88 kN-m |
| MomentY,(My) | = | 38.95 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C13 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 32.692 kN.m |
| My_MinEccen | = | 30.062 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 32.692 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 38.949 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 13.88/1503.07 = 9 mm |
| Actual eccenY | = | My / P = 38.95/1503.07 = 26 mm |
| eccenXMin | = | (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(9,22) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(26,20) = 26 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(26 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 12 nos. + #16 - 4 nos. (2161 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00313 | 355.01 | 8.92 | 346.09 | 139.17 | 177 | 24.63 |
| 2 | 226 | 0.00272 | 350.83 | 8.92 | 341.91 | 77.34 | 125 | 9.67 |
| 3 | 226 | 0.00233 | 339.98 | 8.92 | 331.06 | 74.88 | 75 | 5.62 |
| 4 | 226 | 0.00194 | 325.70 | 8.91 | 316.78 | 71.66 | 25 | 1.79 |
| 5 | 226 | 0.00156 | 299.65 | 8.48 | 291.17 | 65.86 | -25 | -1.65 |
| 6 | 226 | 0.00117 | 233.33 | 7.37 | 225.96 | 51.11 | -75 | -3.83 |
| 7 | 226 | 0.00078 | 155.56 | 5.59 | 149.97 | 33.92 | -125 | -4.24 |
| 8 | 402 | 0.00037 | 74.67 | 3.02 | 71.65 | 28.81 | -177 | -5.10 |
| Total | 542.75 | | 26.89 |
| xux | = | 450 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 450/450 = 0.360 |
| Puc | = | 0.360 x 20.00 x 300 x 450 |
| = | 972.00 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 972.00 + (542.75) | = | 1514.75 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 450/450 |
| = | 0.416 |
| Muc | = | 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 36.74 + (26.89) |
| = | 63.63 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1081 | 0.00288 | 352.85 | 8.92 | 343.93 | 371.69 | 102 | 37.91 |
| 2 | 1081 | 0.00065 | 130.43 | 4.87 | 125.56 | 135.69 | -102 | -13.84 |
| Total | 507.38 | | 24.07 |
| xuy | = | 312 mm Puc = C1.fck.B.D |
| ku | = | 1.039 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 1.001 |
| C1 | = | 0.446 x (1 - C3/6) = 0.372 |
| Puc | = | 0.372 x 20.00 x 450 x 300 |
| = | 1003.24 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 1003.24 + (507.38) | = | 1510.62 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.428 |
| Muc | = | 1003.24 x (0.5 x 300 - 0.428 x 300) = 21.53 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 21.53 + (24.07) |
| = | 45.60 kN-m |
| Pu/Puz | = | 0.805 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.805, | an | = | 2.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((32.69/63.63)2.000) + ((38.95/45.60)2.000) |
| = 0.993 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG5 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 12 nos. + #16 - 4 nos. (2161 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG6 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 278.42 kN |
| MomentX,(Mx) | = | 58.86 kN-m |
| MomentY,(My) | = | 15.28 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C14 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 5.763 kN.m |
| My_MinEccen | = | 5.568 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 58.861 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 15.282 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 58.86/278.42 = 211 mm |
| Actual eccenY | = | My / P = 15.28/278.42 = 55 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(211,21) = 211 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(55,20) = 55 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(211 mm) > 0.05 x 300(15 mm)
and eccenY(55 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00257 | 347.04 | 8.92 | 338.12 | 76.48 | 179 | 13.69 |
| 2 | 226 | 0.00077 | 154.96 | 5.57 | 149.39 | 33.79 | 90 | 3.02 |
| 3 | 226 | -0.00103 | -205.05 | 0.00 | -205.05 | -46.38 | 0 | 0.00 |
| 4 | 226 | -0.00283 | -352.37 | 0.00 | -352.37 | -79.70 | -90 | 7.13 |
| 5 | 226 | -0.00463 | -360.90 | 0.00 | -360.90 | -81.63 | -179 | 14.61 |
| Total | -97.45 | | 38.46 |
| xux | = | 174 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 174/450 = 0.139 |
| Puc | = | 0.139 x 20.00 x 300 x 450 |
| = | 375.89 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 375.89 + (-97.45) | = | 278.44 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 174/450 |
| = | 0.161 |
| Muc | = | 375.89 x (0.5 x 450 - 0.161 x 450) = 57.36 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 57.36 + (38.46) |
| = | 95.82 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00181 | 318.20 | 8.84 | 309.36 | 174.94 | 104 | 18.19 |
| 2 | 565 | -0.00581 | -360.90 | 0.00 | -360.90 | -204.08 | -104 | 21.22 |
| Total | -29.15 | | 39.42 |
| xuy | = | 96 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 96/300 = 0.115 |
| Puc | = | 0.115 x 20.00 x 450 x 300 |
| = | 309.45 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 309.45 + (-29.15) | = | 280.30 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 96/300 |
| = | 0.132 |
| Muc | = | 309.45 x (0.5 x 300 - 0.132 x 300) = 34.12 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 34.12 + (39.42) |
| = | 73.54 kN-m |
| Pu/Puz | = | 0.179 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.179, | an | = | 1.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((58.86/95.82)1.000) + ((15.28/73.54)1.000) |
| = 0.822 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG6 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 558.27 kN |
| MomentX,(Mx) | = | 56.56 kN-m |
| MomentY,(My) | = | 15.38 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C14 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 11.556 kN.m |
| My_MinEccen | = | 11.165 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 56.563 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 15.385 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 56.56/558.27 = 101 mm |
| Actual eccenY | = | My / P = 15.38/558.27 = 28 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(101,21) = 101 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(28,20) = 28 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(101 mm) > 0.05 x 300(15 mm)
and eccenY(28 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00284 | 352.51 | 8.92 | 343.59 | 77.72 | 179 | 13.91 |
| 2 | 226 | 0.00156 | 299.98 | 8.49 | 291.49 | 65.93 | 90 | 5.90 |
| 3 | 226 | 0.00028 | 55.40 | 2.30 | 53.10 | 12.01 | 0 | 0.00 |
| 4 | 226 | -0.00101 | -201.01 | 0.00 | -201.01 | -45.47 | -90 | 4.07 |
| 5 | 226 | -0.00229 | -338.29 | 0.00 | -338.29 | -76.52 | -179 | 13.70 |
| Total | 33.67 | | 37.58 |
| xux | = | 244 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 244/450 = 0.195 |
| Puc | = | 0.195 x 20.00 x 300 x 450 |
| = | 527.77 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 527.77 + (33.67) | = | 561.44 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 244/450 |
| = | 0.226 |
| Muc | = | 527.77 x (0.5 x 450 - 0.226 x 450) = 65.10 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 65.10 + (37.58) |
| = | 102.68 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00255 | 346.30 | 8.92 | 337.38 | 190.78 | 104 | 19.84 |
| 2 | 565 | -0.00177 | -315.32 | 0.00 | -315.32 | -178.31 | -104 | 18.54 |
| Total | 12.47 | | 38.39 |
| xuy | = | 169 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 169/300 = 0.203 |
| Puc | = | 0.203 x 20.00 x 450 x 300 |
| = | 546.75 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 546.75 + (12.47) | = | 559.22 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 169/300 |
| = | 0.234 |
| Muc | = | 546.75 x (0.5 x 300 - 0.234 x 300) = 43.63 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 43.63 + (38.39) |
| = | 82.02 kN-m |
| Pu/Puz | = | 0.359 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.359, | an | = | 1.264 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((56.56/102.68)1.264) + ((15.38/82.02)1.264) |
| = 0.591 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG6 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 833.76 kN |
| MomentX,(Mx) | = | 48.07 kN-m |
| MomentY,(My) | = | 16.91 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C14 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 17.259 kN.m |
| My_MinEccen | = | 16.675 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 48.074 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 16.910 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 48.07/833.76 = 58 mm |
| Actual eccenY | = | My / P = 16.91/833.76 = 20 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(58,21) = 58 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(20,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(58 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00299 | 353.82 | 8.92 | 344.90 | 78.02 | 179 | 13.96 |
| 2 | 226 | 0.00200 | 327.78 | 8.92 | 318.86 | 72.12 | 90 | 6.46 |
| 3 | 226 | 0.00101 | 202.22 | 6.74 | 195.48 | 44.22 | 0 | 0.00 |
| 4 | 226 | 0.00002 | 4.22 | 0.19 | 4.03 | 0.91 | -90 | -0.08 |
| 5 | 226 | -0.00097 | -193.79 | 0.00 | -193.79 | -43.83 | -179 | 7.85 |
| Total | 151.43 | | 28.18 |
| xux | = | 316 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 316/450 = 0.253 |
| Puc | = | 0.253 x 20.00 x 300 x 450 |
| = | 683.44 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 683.44 + (151.43) | = | 834.87 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 316/450 |
| = | 0.293 |
| Muc | = | 683.44 x (0.5 x 450 - 0.293 x 450) = 63.82 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 63.82 + (28.18) |
| = | 92.00 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00276 | 351.81 | 8.92 | 342.89 | 193.90 | 104 | 20.17 |
| 2 | 565 | -0.00058 | -115.71 | 0.00 | -115.71 | -65.43 | -104 | 6.81 |
| Total | 128.47 | | 26.97 |
| xuy | = | 218 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 218/300 = 0.262 |
| Puc | = | 0.262 x 20.00 x 450 x 300 |
| = | 706.22 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 706.22 + (128.47) | = | 834.69 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 218/300 |
| = | 0.302 |
| Muc | = | 706.22 x (0.5 x 300 - 0.302 x 300) = 41.90 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 41.90 + (26.97) |
| = | 68.87 kN-m |
| Pu/Puz | = | 0.536 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.536, | an | = | 1.559 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((48.07/92.00)1.559) + ((16.91/68.87)1.559) |
| = 0.475 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG6 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1148.59 kN |
| MomentX,(Mx) | = | 34.53 kN-m |
| MomentY,(My) | = | 15.09 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C14 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 23.776 kN.m |
| My_MinEccen | = | 22.972 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 34.528 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 22.972 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 34.53/1148.59 = 30 mm |
| Actual eccenY | = | My / P = 15.09/1148.59 = 13 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(30,21) = 30 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(13,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(30 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00311 | 354.88 | 8.92 | 345.96 | 78.25 | 179 | 14.01 |
| 2 | 226 | 0.00236 | 340.85 | 8.92 | 331.93 | 75.08 | 90 | 6.72 |
| 3 | 226 | 0.00160 | 304.02 | 8.57 | 295.45 | 66.83 | 0 | 0.00 |
| 4 | 226 | 0.00085 | 169.32 | 5.95 | 163.36 | 36.95 | -90 | -3.31 |
| 5 | 226 | 0.00009 | 18.30 | 0.80 | 17.50 | 3.96 | -179 | -0.71 |
| Total | 261.07 | | 16.71 |
| xux | = | 415 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 415/450 = 0.332 |
| Puc | = | 0.332 x 20.00 x 300 x 450 |
| = | 896.06 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 896.06 + (261.07) | = | 1157.14 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 415/450 |
| = | 0.384 |
| Muc | = | 896.06 x (0.5 x 450 - 0.384 x 450) = 46.98 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 46.98 + (16.71) |
| = | 63.69 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00293 | 353.31 | 8.92 | 344.39 | 194.75 | 104 | 20.25 |
| 2 | 565 | 0.00037 | 73.05 | 2.96 | 70.09 | 39.63 | -104 | -4.12 |
| Total | 234.38 | | 16.13 |
| xuy | = | 284 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 284/300 = 0.340 |
| Puc | = | 0.340 x 20.00 x 450 x 300 |
| = | 918.84 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 918.84 + (234.38) | = | 1153.22 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 284/300 |
| = | 0.393 |
| Muc | = | 918.84 x (0.5 x 300 - 0.393 x 300) = 29.43 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 29.43 + (16.13) |
| = | 45.56 kN-m |
| Pu/Puz | = | 0.738 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.738, | an | = | 1.896 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((34.53/63.69)1.896) + ((22.97/45.56)1.896) |
| = 0.586 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG6 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3375 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1379.84 kN |
| MomentX,(Mx) | = | 18.18 kN-m |
| MomentY,(My) | = | 6.44 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C14 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 30.012 kN.m |
| My_MinEccen | = | 27.597 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 30.012 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 27.597 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 18.18/1379.84 = 13 mm |
| Actual eccenY | = | My / P = 6.44/1379.84 = 5 mm |
| eccenXMin | = | (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(13,22) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(5,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 14 nos. (1583 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00313 | 355.01 | 8.92 | 346.09 | 78.28 | 179 | 14.01 |
| 2 | 226 | 0.00267 | 349.46 | 8.92 | 340.54 | 77.03 | 119 | 9.19 |
| 3 | 226 | 0.00221 | 335.50 | 8.92 | 326.58 | 73.87 | 60 | 4.41 |
| 4 | 226 | 0.00175 | 314.40 | 8.78 | 305.62 | 69.13 | -0 | -0.00 |
| 5 | 226 | 0.00130 | 259.11 | 7.81 | 251.30 | 56.84 | -60 | -3.39 |
| 6 | 226 | 0.00084 | 167.55 | 5.91 | 161.64 | 36.56 | -119 | -4.36 |
| 7 | 226 | 0.00038 | 75.99 | 3.07 | 72.92 | 16.49 | -179 | -2.95 |
| Total | 408.21 | | 16.91 |
| xux | = | 454 mm Puc = C1.fck.B.D |
| ku | = | 1.008 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 1.112 |
| C1 | = | 0.446 x (1 - C3/6) = 0.363 |
| Puc | = | 0.363 x 20.00 x 300 x 450 |
| = | 980.97 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 980.97 + (408.21) | = | 1389.18 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.419 |
| Muc | = | 980.97 x (0.5 x 450 - 0.419 x 450) = 35.88 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 35.88 + (16.91) |
| = | 52.78 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 792 | 0.00288 | 352.84 | 8.92 | 343.92 | 272.28 | 104 | 28.32 |
| 2 | 792 | 0.00066 | 132.90 | 4.94 | 127.95 | 101.30 | -104 | -10.54 |
| Total | 373.58 | | 17.78 |
| xuy | = | 316 mm Puc = C1.fck.B.D |
| ku | = | 1.055 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 0.952 |
| C1 | = | 0.446 x (1 - C3/6) = 0.375 |
| Puc | = | 0.375 x 20.00 x 450 x 300 |
| = | 1013.15 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 1013.15 + (373.58) | = | 1386.72 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.433 |
| Muc | = | 1013.15 x (0.5 x 300 - 0.433 x 300) = 20.47 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 20.47 + (17.78) |
| = | 38.25 kN-m |
| Pu/Puz | = | 0.815 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.815, | an | = | 2.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((30.01/52.78)2.000) + ((27.60/38.25)2.000) |
| = 0.844 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG6 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 14 nos. (1583 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG7 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3450 mm |
From analysis results,loads on column
| Axial load,(P) | = | 387.26 kN |
| MomentX,(Mx) | = | 127.59 kN-m |
| MomentY,(My) | = | 14.59 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C15 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 8.481 kN.m |
| My_MinEccen | = | 7.745 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 127.586 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 14.588 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 127.59/387.26 = 329 mm |
| Actual eccenY | = | My / P = 14.59/387.26 = 38 mm |
| eccenXMin | = | (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(329,22) = 329 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(38,20) = 38 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(329 mm) > 0.05 x 300(15 mm)
and eccenY(38 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. + #16 - 4 nos. (2614 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00270 | 350.35 | 8.92 | 341.43 | 137.30 | 177 | 24.30 |
| 2 | 226 | 0.00203 | 328.66 | 8.92 | 319.74 | 72.32 | 136 | 9.84 |
| 3 | 226 | 0.00138 | 275.97 | 8.06 | 267.90 | 60.60 | 97 | 5.89 |
| 4 | 226 | 0.00073 | 146.91 | 5.35 | 141.56 | 32.02 | 58 | 1.87 |
| 5 | 226 | 0.00009 | 17.86 | 0.78 | 17.08 | 3.86 | 19 | 0.08 |
| 6 | 226 | -0.00056 | -111.19 | 0.00 | -111.19 | -25.15 | -19 | 0.49 |
| 7 | 226 | -0.00120 | -240.25 | 0.00 | -240.25 | -54.34 | -58 | 3.17 |
| 8 | 226 | -0.00185 | -320.21 | 0.00 | -320.21 | -72.43 | -97 | 7.04 |
| 9 | 226 | -0.00249 | -344.90 | 0.00 | -344.90 | -78.02 | -136 | 10.62 |
| 10 | 402 | -0.00317 | -355.39 | 0.00 | -355.39 | -142.91 | -177 | 25.30 |
| Total | -66.75 | | 88.59 |
| xux | = | 211 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 211/450 = 0.169 |
| Puc | = | 0.169 x 20.00 x 300 x 450 |
| = | 455.63 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 455.63 + (-66.75) | = | 388.88 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 211/450 |
| = | 0.195 |
| Muc | = | 455.63 x (0.5 x 450 - 0.195 x 450) = 62.53 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 62.53 + (88.59) |
| = | 151.13 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1307 | 0.00222 | 335.82 | 8.92 | 326.90 | 427.23 | 102 | 43.58 |
| 2 | 1307 | -0.00322 | -355.83 | 0.00 | -355.83 | -465.03 | -102 | 47.43 |
| Total | -37.80 | | 91.01 |
| xuy | = | 131 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 131/300 = 0.157 |
| Puc | = | 0.157 x 20.00 x 450 x 300 |
| = | 425.25 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 425.25 + (-37.80) | = | 387.45 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 131/300 |
| = | 0.182 |
| Muc | = | 425.25 x (0.5 x 300 - 0.182 x 300) = 40.57 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 40.57 + (91.01) |
| = | 131.58 kN-m |
| Pu/Puz | = | 0.193 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.193, | an | = | 1.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((127.59/151.13)1.000) + ((14.59/131.58)1.000) |
| = 0.955 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG7 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. + #16 - 4 nos. (2614 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3450 mm |
From analysis results,loads on column
| Axial load,(P) | = | 790.58 kN |
| MomentX,(Mx) | = | 112.81 kN-m |
| MomentY,(My) | = | 14.18 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C15 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 17.314 kN.m |
| My_MinEccen | = | 15.812 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 112.812 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 15.812 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 112.81/790.58 = 143 mm |
| Actual eccenY | = | My / P = 14.18/790.58 = 18 mm |
| eccenXMin | = | (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(143,22) = 143 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(18,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(143 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 2 nos. + #16 - 8 nos. (1835 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00292 | 353.21 | 8.92 | 344.29 | 138.45 | 177 | 24.50 |
| 2 | 402 | 0.00184 | 319.85 | 8.86 | 310.99 | 125.06 | 88 | 10.94 |
| 3 | 226 | 0.00078 | 156.97 | 5.63 | 151.34 | 34.23 | 0 | 0.00 |
| 4 | 402 | -0.00027 | -54.21 | 0.00 | -54.21 | -21.80 | -88 | 1.91 |
| 5 | 402 | -0.00135 | -270.21 | 0.00 | -270.21 | -108.66 | -177 | 19.23 |
| Total | 167.28 | | 56.59 |
| xux | = | 290 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 290/450 = 0.232 |
| Puc | = | 0.232 x 20.00 x 300 x 450 |
| = | 626.48 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 626.48 + (167.28) | = | 793.76 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 290/450 |
| = | 0.268 |
| Muc | = | 626.48 x (0.5 x 450 - 0.268 x 450) = 65.37 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 65.37 + (56.59) |
| = | 121.96 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 917 | 0.00266 | 349.27 | 8.92 | 340.35 | 312.22 | 102 | 31.85 |
| 2 | 917 | -0.00090 | -180.28 | 0.00 | -180.28 | -165.38 | -102 | 16.87 |
| Total | 146.84 | | 48.72 |
| xuy | = | 200 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 200/300 = 0.240 |
| Puc | = | 0.240 x 20.00 x 450 x 300 |
| = | 649.27 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 649.27 + (146.84) | = | 796.10 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 200/300 |
| = | 0.278 |
| Muc | = | 649.27 x (0.5 x 300 - 0.278 x 300) = 43.27 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 43.27 + (48.72) |
| = | 91.98 kN-m |
| Pu/Puz | = | 0.447 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.447, | an | = | 1.411 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((112.81/121.96)1.411) + ((15.81/91.98)1.411) |
| = 0.979 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG7 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 2 nos. + #16 - 8 nos. (1835 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3150 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1200.61 kN |
| MomentX,(Mx) | = | 97.34 kN-m |
| MomentY,(My) | = | 9.59 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C15 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3150 x 1.000 ) / 450 = 7.00 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3150 x 1.000 ) / 300 = 10.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 25.573 kN.m |
| My_MinEccen | = | 24.012 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 97.343 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 24.012 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 97.34/1200.61 = 81 mm |
| Actual eccenY | = | My / P = 9.59/1200.61 = 8 mm |
| eccenXMin | = | (L/500) + (D/30) = (3150/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (3150/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(81,21) = 81 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(8,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(81 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. + #16 - 4 nos. (2614 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00302 | 354.05 | 8.92 | 345.13 | 138.79 | 177 | 24.57 |
| 2 | 226 | 0.00261 | 347.84 | 8.92 | 338.92 | 76.66 | 136 | 10.43 |
| 3 | 226 | 0.00222 | 335.64 | 8.92 | 326.72 | 73.90 | 97 | 7.18 |
| 4 | 226 | 0.00182 | 318.81 | 8.85 | 309.96 | 70.11 | 58 | 4.09 |
| 5 | 226 | 0.00143 | 286.58 | 8.20 | 278.38 | 62.97 | 19 | 1.22 |
| 6 | 226 | 0.00104 | 208.37 | 6.87 | 201.49 | 45.58 | -19 | -0.89 |
| 7 | 226 | 0.00065 | 130.15 | 4.86 | 125.29 | 28.34 | -58 | -1.65 |
| 8 | 226 | 0.00026 | 51.94 | 2.17 | 49.77 | 11.26 | -97 | -1.09 |
| 9 | 226 | -0.00013 | -26.28 | 0.00 | -26.28 | -5.94 | -136 | 0.81 |
| 10 | 402 | -0.00054 | -108.51 | 0.00 | -108.51 | -43.64 | -177 | 7.72 |
| Total | 458.03 | | 52.40 |
| xux | = | 348 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 348/450 = 0.278 |
| Puc | = | 0.278 x 20.00 x 300 x 450 |
| = | 751.78 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 751.78 + (458.03) | = | 1209.81 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 348/450 |
| = | 0.322 |
| Muc | = | 751.78 x (0.5 x 450 - 0.322 x 450) = 60.30 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 60.30 + (52.40) |
| = | 112.70 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1307 | 0.00281 | 352.24 | 8.92 | 343.32 | 448.69 | 102 | 45.77 |
| 2 | 1307 | -0.00012 | -23.69 | 0.00 | -23.69 | -30.96 | -102 | 3.16 |
| Total | 417.73 | | 48.92 |
| xuy | = | 244 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 244/300 = 0.293 |
| Puc | = | 0.293 x 20.00 x 450 x 300 |
| = | 789.75 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 789.75 + (417.73) | = | 1207.48 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 244/300 |
| = | 0.338 |
| Muc | = | 789.75 x (0.5 x 300 - 0.338 x 300) = 38.38 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 38.38 + (48.92) |
| = | 87.31 kN-m |
| Pu/Puz | = | 0.599 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.599, | an | = | 1.665 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((97.34/112.70)1.665) + ((24.01/87.31)1.665) |
| = 0.900 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG7 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. + #16 - 4 nos. (2614 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3150 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1779.89 kN |
| MomentX,(Mx) | = | 92.72 kN-m |
| MomentY,(My) | = | 5.20 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C15 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3150 x 1.000 ) / 450 = 7.00 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3150 x 1.000 ) / 300 = 10.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 37.912 kN.m |
| My_MinEccen | = | 35.598 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 92.722 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 35.598 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 92.72/1779.89 = 52 mm |
| Actual eccenY | = | My / P = 5.20/1779.89 = 3 mm |
| eccenXMin | = | (L/500) + (D/30) = (3150/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (3150/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(52,21) = 52 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(52 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 628 | 0.00305 | 354.32 | 8.92 | 345.40 | 217.02 | 175 | 37.98 |
| 2 | 628 | 0.00252 | 345.62 | 8.92 | 336.70 | 211.55 | 117 | 24.68 |
| 3 | 628 | 0.00199 | 327.43 | 8.92 | 318.51 | 200.13 | 58 | 11.67 |
| 4 | 628 | 0.00146 | 290.94 | 8.28 | 282.66 | 177.60 | 0 | 0.00 |
| 5 | 628 | 0.00094 | 187.14 | 6.39 | 180.74 | 113.56 | -58 | -6.62 |
| 6 | 628 | 0.00041 | 81.55 | 3.27 | 78.28 | 49.19 | -117 | -5.74 |
| 7 | 628 | -0.00012 | -24.04 | 0.00 | -24.04 | -15.11 | -175 | 2.64 |
| Total | 953.94 | | 64.61 |
| xux | = | 387 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 387/450 = 0.309 |
| Puc | = | 0.309 x 20.00 x 300 x 450 |
| = | 835.31 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 835.31 + (953.94) | = | 1789.26 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 387/450 |
| = | 0.358 |
| Muc | = | 835.31 x (0.5 x 450 - 0.358 x 450) = 53.56 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 53.56 + (64.61) |
| = | 118.18 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 2199 | 0.00287 | 352.74 | 8.92 | 343.82 | 756.10 | 100 | 75.61 |
| 2 | 2199 | 0.00034 | 67.23 | 2.75 | 64.49 | 141.81 | -100 | -14.18 |
| Total | 897.91 | | 61.43 |
| xuy | = | 277 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 277/300 = 0.332 |
| Puc | = | 0.332 x 20.00 x 450 x 300 |
| = | 896.06 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 896.06 + (897.91) | = | 1793.97 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 277/300 |
| = | 0.384 |
| Muc | = | 896.06 x (0.5 x 300 - 0.384 x 300) = 31.32 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 31.32 + (61.43) |
| = | 92.75 kN-m |
| Pu/Puz | = | 0.700 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.700, | an | = | 1.833 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((92.72/118.18)1.833) + ((35.60/92.75)1.833) |
| = 0.814 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 20 = 320 mm
iii) 300 mm
Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties 8 @ 300 mm c/c
SUMMARY :
CG7 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3675 mm |
From analysis results,loads on column
| Axial load,(P) | = | 2208.46 kN |
| MomentX,(Mx) | = | 76.81 kN-m |
| MomentY,(My) | = | 5.73 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C15 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3675 x 1.000 ) / 450 = 8.17 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3675 x 1.000 ) / 300 = 12.25 > 12.00, |
column is slender in this direction.| Puz | = | (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast |
| = | 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 5890.49 = 2995399.658 N |
| k1 | = | 0.190000 |
| k2 | = | 0.012000 MPa |
| Pub | = | (k1 x fck x Ag) + (k2 x Ast x 100) |
| = | (0.190000 x 20.00 x 135000.01) + (0.01 x 5890.49 x 100) = 520068.604 N |
| k | = | (Puz - Pu) / (Puz - Pub) <= 1 |
| = | (2995399.658 - 2208460.449) / (2995399.658 - 520068.604) = 0.317913 |
| MuaddY | = | Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k |
| = | 2208460.45 x (300.00/2000) x (3675.00/300.00) x (3675.00/300.00) x 0.317913 |
| = | 15803779.602 N.mm = 15.804 kN.m |
| Mx_MinEccen | = | 49.359 kN.m |
| My_MinEccen | = | 44.169 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 76.806 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 59.973 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 76.81/2208.46 = 35 mm |
| Actual eccenY | = | My / P = 5.73/2208.46 = 3 mm |
| eccenXMin | = | (L/500) + (D/30) = (3675/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3675/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(35,22) = 35 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(35 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 12 nos. (5890 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 982 | 0.00305 | 354.30 | 8.92 | 345.38 | 339.07 | 173 | 58.49 |
| 2 | 982 | 0.00245 | 343.78 | 8.92 | 334.86 | 328.75 | 104 | 34.03 |
| 3 | 982 | 0.00185 | 320.48 | 8.87 | 311.61 | 305.93 | 35 | 10.55 |
| 4 | 982 | 0.00125 | 250.70 | 7.68 | 243.02 | 238.59 | -35 | -8.23 |
| 5 | 982 | 0.00066 | 131.23 | 4.89 | 126.34 | 124.04 | -104 | -12.84 |
| 6 | 982 | 0.00006 | 11.77 | 0.52 | 11.25 | 11.05 | -173 | -1.91 |
| Total | 1347.42 | | 80.10 |
| xux | = | 404 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 404/450 = 0.323 |
| Puc | = | 0.323 x 20.00 x 300 x 450 |
| = | 873.28 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 873.28 + (1347.42) | = | 2220.70 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 404/450 |
| = | 0.374 |
| Muc | = | 873.28 x (0.5 x 450 - 0.374 x 450) = 49.61 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 49.61 + (80.10) |
| = | 129.71 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 2945 | 0.00286 | 352.70 | 8.92 | 343.78 | 1012.51 | 98 | 98.72 |
| 2 | 2945 | 0.00050 | 99.02 | 3.87 | 95.15 | 280.25 | -98 | -27.32 |
| Total | 1292.76 | | 71.39 |
| xuy | = | 288 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 288/300 = 0.346 |
| Puc | = | 0.346 x 20.00 x 450 x 300 |
| = | 934.03 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 934.03 + (1292.76) | = | 2226.79 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 288/300 |
| = | 0.400 |
| Muc | = | 934.03 x (0.5 x 300 - 0.400 x 300) = 28.09 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 28.09 + (71.39) |
| = | 99.49 kN-m |
| Pu/Puz | = | 0.737 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.737, | an | = | 1.895 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((76.81/129.71)1.895) + ((59.97/99.49)1.895) |
| = 0.754 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 25 = 400 mm
iii) 300 mm
Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties 8 @ 300 mm c/c
SUMMARY :
CG7 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 12 nos. (5890 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG8 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3450 mm |
From analysis results,loads on column
| Axial load,(P) | = | 261.01 kN |
| MomentX,(Mx) | = | 152.50 kN-m |
| MomentY,(My) | = | 12.19 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C16 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 5.716 kN.m |
| My_MinEccen | = | 5.220 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 152.503 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 12.187 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 152.50/261.01 = 584 mm |
| Actual eccenY | = | My / P = 12.19/261.01 = 47 mm |
| eccenXMin | = | (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(584,22) = 584 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(47,20) = 47 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(584 mm) > 0.05 x 300(15 mm)
and eccenY(47 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 18 nos. (3619 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00266 | 349.27 | 8.92 | 340.35 | 136.86 | 177 | 24.22 |
| 2 | 402 | 0.00189 | 322.85 | 8.89 | 313.96 | 126.25 | 133 | 16.76 |
| 3 | 402 | 0.00112 | 223.18 | 7.18 | 216.00 | 86.86 | 89 | 7.69 |
| 4 | 402 | 0.00034 | 68.61 | 2.80 | 65.81 | 26.46 | 44 | 1.17 |
| 5 | 402 | -0.00043 | -85.96 | 0.00 | -85.96 | -34.57 | 0 | 0.00 |
| 6 | 402 | -0.00120 | -240.54 | 0.00 | -240.54 | -96.73 | -44 | 4.28 |
| 7 | 402 | -0.00198 | -326.84 | 0.00 | -326.84 | -131.43 | -89 | 11.63 |
| 8 | 402 | -0.00275 | -351.50 | 0.00 | -351.50 | -141.35 | -133 | 18.76 |
| 9 | 402 | -0.00352 | -358.46 | 0.00 | -358.46 | -144.15 | -177 | 25.51 |
| Total | -171.78 | | 110.03 |
| xux | = | 200 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 200/450 = 0.160 |
| Puc | = | 0.160 x 20.00 x 300 x 450 |
| = | 432.84 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 432.84 + (-171.78) | = | 261.06 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 200/450 |
| = | 0.185 |
| Muc | = | 432.84 x (0.5 x 450 - 0.185 x 450) = 61.31 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 61.31 + (110.03) |
| = | 171.34 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1810 | 0.00192 | 324.54 | 8.90 | 315.64 | 571.17 | 102 | 58.26 |
| 2 | 1810 | -0.00482 | -360.90 | 0.00 | -360.90 | -653.07 | -102 | 66.61 |
| Total | -81.90 | | 124.87 |
| xuy | = | 106 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 106/300 = 0.127 |
| Puc | = | 0.127 x 20.00 x 450 x 300 |
| = | 343.62 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 343.62 + (-81.90) | = | 261.72 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 106/300 |
| = | 0.147 |
| Muc | = | 343.62 x (0.5 x 300 - 0.147 x 300) = 36.38 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 36.38 + (124.87) |
| = | 161.26 kN-m |
| Pu/Puz | = | 0.113 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.113, | an | = | 1.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((152.50/171.34)1.000) + ((12.19/161.26)1.000) |
| = 0.966 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG8 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 18 nos. (3619 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3450 mm |
From analysis results,loads on column
| Axial load,(P) | = | 541.65 kN |
| MomentX,(Mx) | = | 98.60 kN-m |
| MomentY,(My) | = | 9.32 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C16 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 11.862 kN.m |
| My_MinEccen | = | 10.833 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 98.603 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 10.833 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 98.60/541.65 = 182 mm |
| Actual eccenY | = | My / P = 9.32/541.65 = 17 mm |
| eccenXMin | = | (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(182,22) = 182 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(17,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(182 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 6 nos. (1206 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00281 | 352.22 | 8.92 | 343.30 | 138.05 | 177 | 24.43 |
| 2 | 402 | 0.00025 | 50.72 | 2.12 | 48.61 | 19.55 | 0 | 0.00 |
| 3 | 402 | -0.00230 | -338.77 | 0.00 | -338.77 | -136.23 | -177 | 24.11 |
| Total | 21.37 | | 48.55 |
| xux | = | 243 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 243/450 = 0.194 |
| Puc | = | 0.194 x 20.00 x 300 x 450 |
| = | 523.97 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 523.97 + (21.37) | = | 545.34 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 243/450 |
| = | 0.224 |
| Muc | = | 523.97 x (0.5 x 450 - 0.224 x 450) = 65.02 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 65.02 + (48.55) |
| = | 113.56 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 603 | 0.00248 | 344.68 | 8.92 | 335.76 | 202.53 | 102 | 20.66 |
| 2 | 603 | -0.00184 | -319.67 | 0.00 | -319.67 | -192.82 | -102 | 19.67 |
| Total | 9.71 | | 40.33 |
| xuy | = | 165 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 165/300 = 0.198 |
| Puc | = | 0.198 x 20.00 x 450 x 300 |
| = | 535.36 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 535.36 + (9.71) | = | 545.06 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 165/300 |
| = | 0.229 |
| Muc | = | 535.36 x (0.5 x 300 - 0.229 x 300) = 43.50 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 43.50 + (40.33) |
| = | 83.83 kN-m |
| Pu/Puz | = | 0.343 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.343, | an | = | 1.238 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((98.60/113.56)1.238) + ((10.83/83.83)1.238) |
| = 0.919 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG8 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 6 nos. (1206 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3150 mm |
From analysis results,loads on column
| Axial load,(P) | = | 815.23 kN |
| MomentX,(Mx) | = | 90.57 kN-m |
| MomentY,(My) | = | 6.11 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C16 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3150 x 1.000 ) / 450 = 7.00 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3150 x 1.000 ) / 300 = 10.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 17.365 kN.m |
| My_MinEccen | = | 16.305 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 90.567 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 16.305 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 90.57/815.23 = 111 mm |
| Actual eccenY | = | My / P = 6.11/815.23 = 7 mm |
| eccenXMin | = | (L/500) + (D/30) = (3150/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (3150/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(111,21) = 111 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(7,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(111 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 6 nos. (1206 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00297 | 353.60 | 8.92 | 344.68 | 138.61 | 177 | 24.53 |
| 2 | 402 | 0.00100 | 199.44 | 6.68 | 192.76 | 77.51 | 0 | 0.00 |
| 3 | 402 | -0.00097 | -194.33 | 0.00 | -194.33 | -78.15 | -177 | 13.83 |
| Total | 137.97 | | 38.36 |
| xux | = | 315 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 315/450 = 0.252 |
| Puc | = | 0.252 x 20.00 x 300 x 450 |
| = | 679.64 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 679.64 + (137.97) | = | 817.62 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 315/450 |
| = | 0.291 |
| Muc | = | 679.64 x (0.5 x 450 - 0.291 x 450) = 63.96 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 63.96 + (38.36) |
| = | 102.32 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 603 | 0.00271 | 350.57 | 8.92 | 341.65 | 206.08 | 102 | 21.02 |
| 2 | 603 | -0.00064 | -127.08 | 0.00 | -127.08 | -76.65 | -102 | 7.82 |
| Total | 129.43 | | 28.84 |
| xuy | = | 213 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 213/300 = 0.256 |
| Puc | = | 0.256 x 20.00 x 450 x 300 |
| = | 691.03 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 691.03 + (129.43) | = | 820.46 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 213/300 |
| = | 0.296 |
| Muc | = | 691.03 x (0.5 x 300 - 0.296 x 300) = 42.34 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 42.34 + (28.84) |
| = | 71.18 kN-m |
| Pu/Puz | = | 0.516 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.516, | an | = | 1.527 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((90.57/102.32)1.527) + ((16.30/71.18)1.527) |
| = 0.935 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG8 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 6 nos. (1206 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3150 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1562.72 kN |
| MomentX,(Mx) | = | 91.77 kN-m |
| MomentY,(My) | = | 5.99 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C16 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3150 x 1.000 ) / 450 = 7.00 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3150 x 1.000 ) / 300 = 10.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 33.286 kN.m |
| My_MinEccen | = | 31.254 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 91.773 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 31.254 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 91.77/1562.72 = 59 mm |
| Actual eccenY | = | My / P = 5.99/1562.72 = 4 mm |
| eccenXMin | = | (L/500) + (D/30) = (3150/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (3150/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(59,21) = 59 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(4,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(59 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 18 nos. (3619 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00305 | 354.37 | 8.92 | 345.45 | 138.91 | 177 | 24.59 |
| 2 | 402 | 0.00264 | 348.76 | 8.92 | 339.84 | 136.66 | 133 | 18.14 |
| 3 | 402 | 0.00223 | 336.19 | 8.92 | 327.27 | 131.60 | 89 | 11.65 |
| 4 | 402 | 0.00182 | 318.45 | 8.85 | 309.60 | 124.50 | 44 | 5.51 |
| 5 | 402 | 0.00141 | 281.31 | 8.13 | 273.17 | 109.85 | 0 | 0.00 |
| 6 | 402 | 0.00099 | 198.97 | 6.67 | 192.30 | 77.33 | -44 | -3.42 |
| 7 | 402 | 0.00058 | 116.62 | 4.44 | 112.18 | 45.11 | -89 | -3.99 |
| 8 | 402 | 0.00017 | 34.28 | 1.46 | 32.82 | 13.20 | -133 | -1.75 |
| 9 | 402 | -0.00024 | -48.06 | 0.00 | -48.06 | -19.33 | -177 | 3.42 |
| Total | 757.83 | | 54.14 |
| xux | = | 376 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 376/450 = 0.301 |
| Puc | = | 0.301 x 20.00 x 300 x 450 |
| = | 812.53 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 812.53 + (757.83) | = | 1570.36 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 376/450 |
| = | 0.348 |
| Muc | = | 812.53 x (0.5 x 450 - 0.348 x 450) = 55.67 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 55.67 + (54.14) |
| = | 109.81 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1810 | 0.00288 | 352.82 | 8.92 | 343.90 | 622.31 | 102 | 63.48 |
| 2 | 1810 | 0.00023 | 45.53 | 1.92 | 43.62 | 78.92 | -102 | -8.05 |
| Total | 701.23 | | 55.43 |
| xuy | = | 270 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 270/300 = 0.323 |
| Puc | = | 0.323 x 20.00 x 450 x 300 |
| = | 873.28 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 873.28 + (701.23) | = | 1574.51 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 270/300 |
| = | 0.374 |
| Muc | = | 873.28 x (0.5 x 300 - 0.374 x 300) = 33.08 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 33.08 + (55.43) |
| = | 88.50 kN-m |
| Pu/Puz | = | 0.677 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.677, | an | = | 1.795 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((91.77/109.81)1.795) + ((31.25/88.50)1.795) |
| = 0.879 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG8 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 18 nos. (3619 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3675 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1819.48 kN |
| MomentX,(Mx) | = | 77.39 kN-m |
| MomentY,(My) | = | 5.52 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C16 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3675 x 1.000 ) / 450 = 8.17 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3675 x 1.000 ) / 300 = 12.25 > 12.00, |
column is slender in this direction.| Puz | = | (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast |
| = | 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 4398.23 = 2544364.990 N |
| k1 | = | 0.192000 |
| k2 | = | 0.023333 MPa |
| Pub | = | (k1 x fck x Ag) + (k2 x Ast x 100) |
| = | (0.192000 x 20.00 x 135000.01) + (0.02 x 4398.23 x 100) = 528662.598 N |
| k | = | (Puz - Pu) / (Puz - Pub) <= 1 |
| = | (2544364.990 - 1819484.985) / (2544364.990 - 528662.598) = 0.359617 |
| MuaddY | = | Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k |
| = | 1819484.99 x (300.00/2000) x (3675.00/300.00) x (3675.00/300.00) x 0.359617 |
| = | 14728266.716 N.mm = 14.728 kN.m |
| Mx_MinEccen | = | 40.665 kN.m |
| My_MinEccen | = | 36.390 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 77.386 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 51.118 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 77.39/1819.48 = 43 mm |
| Actual eccenY | = | My / P = 5.52/1819.48 = 3 mm |
| eccenXMin | = | (L/500) + (D/30) = (3675/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3675/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(43,22) = 43 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(43 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 628 | 0.00306 | 354.39 | 8.92 | 345.47 | 217.06 | 175 | 37.99 |
| 2 | 628 | 0.00254 | 346.07 | 8.92 | 337.15 | 211.84 | 117 | 24.71 |
| 3 | 628 | 0.00202 | 328.42 | 8.92 | 319.50 | 200.75 | 58 | 11.71 |
| 4 | 628 | 0.00150 | 294.38 | 8.36 | 286.02 | 179.71 | 0 | 0.00 |
| 5 | 628 | 0.00098 | 196.30 | 6.61 | 189.69 | 119.19 | -58 | -6.95 |
| 6 | 628 | 0.00046 | 92.59 | 3.65 | 88.94 | 55.88 | -117 | -6.52 |
| 7 | 628 | -0.00006 | -11.11 | 0.00 | -11.11 | -6.98 | -175 | 1.22 |
| Total | 977.45 | | 62.16 |
| xux | = | 394 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 394/450 = 0.315 |
| Puc | = | 0.315 x 20.00 x 300 x 450 |
| = | 850.50 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 850.50 + (977.45) | = | 1827.95 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 394/450 |
| = | 0.364 |
| Muc | = | 850.50 x (0.5 x 450 - 0.364 x 450) = 52.05 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 52.05 + (62.16) |
| = | 114.21 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 2199 | 0.00288 | 352.83 | 8.92 | 343.91 | 756.30 | 100 | 75.63 |
| 2 | 2199 | 0.00039 | 77.78 | 3.13 | 74.65 | 164.16 | -100 | -16.42 |
| Total | 920.45 | | 59.21 |
| xuy | = | 281 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 281/300 = 0.338 |
| Puc | = | 0.338 x 20.00 x 450 x 300 |
| = | 911.25 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 911.25 + (920.45) | = | 1831.70 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 281/300 |
| = | 0.390 |
| Muc | = | 911.25 x (0.5 x 300 - 0.390 x 300) = 30.07 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 30.07 + (59.21) |
| = | 89.29 kN-m |
| Pu/Puz | = | 0.715 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.715, | an | = | 1.859 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((77.39/114.21)1.859) + ((51.12/89.29)1.859) |
| = 0.840 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 20 = 320 mm
iii) 300 mm
Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties 8 @ 300 mm c/c
SUMMARY :
CG8 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG9 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 305.84 kN |
| MomentX,(Mx) | = | 3.88 kN-m |
| MomentY,(My) | = | 59.81 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C17 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 6.331 kN.m |
| My_MinEccen | = | 6.117 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 6.331 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 59.814 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 3.88/305.84 = 13 mm |
| Actual eccenY | = | My / P = 59.81/305.84 = 196 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(13,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(196,20) = 196 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(196 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00261 | 347.96 | 8.92 | 339.04 | 76.69 | 179 | 13.73 |
| 2 | 226 | 0.00088 | 176.13 | 6.13 | 170.00 | 38.45 | 90 | 3.44 |
| 3 | 226 | -0.00085 | -169.90 | 0.00 | -169.90 | -38.43 | 0 | 0.00 |
| 4 | 226 | -0.00258 | -347.16 | 0.00 | -347.16 | -78.53 | -90 | 7.03 |
| 5 | 226 | -0.00431 | -360.90 | 0.00 | -360.90 | -81.63 | -179 | 14.61 |
| Total | -83.45 | | 38.81 |
| xux | = | 181 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 181/450 = 0.145 |
| Puc | = | 0.145 x 20.00 x 300 x 450 |
| = | 391.08 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 391.08 + (-83.45) | = | 307.63 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 181/450 |
| = | 0.167 |
| Muc | = | 391.08 x (0.5 x 450 - 0.167 x 450) = 58.54 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 58.54 + (38.81) |
| = | 97.35 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00193 | 325.16 | 8.91 | 316.25 | 178.84 | 104 | 18.60 |
| 2 | 565 | -0.00517 | -360.90 | 0.00 | -360.90 | -204.08 | -104 | 21.22 |
| Total | -25.25 | | 39.82 |
| xuy | = | 103 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 103/300 = 0.123 |
| Puc | = | 0.123 x 20.00 x 450 x 300 |
| = | 332.23 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 332.23 + (-25.25) | = | 306.98 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 103/300 |
| = | 0.142 |
| Muc | = | 332.23 x (0.5 x 300 - 0.142 x 300) = 35.66 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 35.66 + (39.82) |
| = | 75.49 kN-m |
| Pu/Puz | = | 0.196 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.196, | an | = | 1.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((6.33/97.35)1.000) + ((59.81/75.49)1.000) |
| = 0.857 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG9 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 615.31 kN |
| MomentX,(Mx) | = | 3.32 kN-m |
| MomentY,(My) | = | 54.52 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C17 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 12.737 kN.m |
| My_MinEccen | = | 12.306 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 12.737 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 54.518 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 3.32/615.31 = 5 mm |
| Actual eccenY | = | My / P = 54.52/615.31 = 89 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(5,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(89,20) = 89 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(89 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00288 | 352.84 | 8.92 | 343.92 | 77.79 | 179 | 13.92 |
| 2 | 226 | 0.00167 | 309.25 | 8.68 | 300.57 | 67.99 | 90 | 6.08 |
| 3 | 226 | 0.00046 | 92.54 | 3.65 | 88.89 | 20.11 | 0 | 0.00 |
| 4 | 226 | -0.00075 | -149.09 | 0.00 | -149.09 | -33.72 | -90 | 3.02 |
| 5 | 226 | -0.00195 | -326.04 | 0.00 | -326.04 | -73.75 | -179 | 13.20 |
| Total | 58.42 | | 36.23 |
| xux | = | 259 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 259/450 = 0.207 |
| Puc | = | 0.207 x 20.00 x 300 x 450 |
| = | 560.04 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 560.04 + (58.42) | = | 618.46 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 259/450 |
| = | 0.240 |
| Muc | = | 560.04 x (0.5 x 450 - 0.240 x 450) = 65.60 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 65.60 + (36.23) |
| = | 101.83 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00261 | 347.96 | 8.92 | 339.04 | 191.72 | 104 | 19.94 |
| 2 | 565 | -0.00141 | -282.02 | 0.00 | -282.02 | -159.48 | -104 | 16.59 |
| Total | 32.24 | | 36.53 |
| xuy | = | 181 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 181/300 = 0.217 |
| Puc | = | 0.217 x 20.00 x 450 x 300 |
| = | 586.62 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 586.62 + (32.24) | = | 618.86 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 181/300 |
| = | 0.251 |
| Muc | = | 586.62 x (0.5 x 300 - 0.251 x 300) = 43.81 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 43.81 + (36.53) |
| = | 80.33 kN-m |
| Pu/Puz | = | 0.395 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.395, | an | = | 1.325 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((12.74/101.83)1.325) + ((54.52/80.33)1.325) |
| = 0.662 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG9 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 933.62 kN |
| MomentX,(Mx) | = | 3.66 kN-m |
| MomentY,(My) | = | 53.03 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C17 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 19.326 kN.m |
| My_MinEccen | = | 18.672 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 19.326 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 53.028 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 3.66/933.62 = 4 mm |
| Actual eccenY | = | My / P = 53.03/933.62 = 57 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(4,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(57,20) = 57 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(57 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00303 | 354.19 | 8.92 | 345.27 | 78.10 | 179 | 13.98 |
| 2 | 226 | 0.00212 | 332.28 | 8.92 | 323.36 | 73.14 | 90 | 6.55 |
| 3 | 226 | 0.00121 | 242.86 | 7.54 | 235.31 | 53.23 | 0 | 0.00 |
| 4 | 226 | 0.00031 | 61.02 | 2.51 | 58.50 | 13.23 | -90 | -1.18 |
| 5 | 226 | -0.00060 | -120.83 | 0.00 | -120.83 | -27.33 | -179 | 4.89 |
| Total | 190.37 | | 24.23 |
| xux | = | 345 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 345/450 = 0.276 |
| Puc | = | 0.276 x 20.00 x 300 x 450 |
| = | 744.19 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 744.19 + (190.37) | = | 934.56 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 345/450 |
| = | 0.319 |
| Muc | = | 744.19 x (0.5 x 450 - 0.319 x 450) = 60.78 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 60.78 + (24.23) |
| = | 85.01 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00282 | 352.35 | 8.92 | 343.43 | 194.21 | 104 | 20.20 |
| 2 | 565 | -0.00024 | -47.40 | 0.00 | -47.40 | -26.81 | -104 | 2.79 |
| Total | 167.40 | | 22.99 |
| xuy | = | 238 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 238/300 = 0.285 |
| Puc | = | 0.285 x 20.00 x 450 x 300 |
| = | 770.77 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 770.77 + (167.40) | = | 938.17 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 238/300 |
| = | 0.330 |
| Muc | = | 770.77 x (0.5 x 300 - 0.330 x 300) = 39.34 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 39.34 + (22.99) |
| = | 62.32 kN-m |
| Pu/Puz | = | 0.600 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.600, | an | = | 1.666 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((19.33/85.01)1.666) + ((53.03/62.32)1.666) |
| = 0.849 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG9 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1265.52 kN |
| MomentX,(Mx) | = | 3.97 kN-m |
| MomentY,(My) | = | 51.12 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C17 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 26.196 kN.m |
| My_MinEccen | = | 25.310 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 26.196 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 51.117 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 3.97/1265.52 = 3 mm |
| Actual eccenY | = | My / P = 51.12/1265.52 = 40 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(40,20) = 40 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(40 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. (1810 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00310 | 354.76 | 8.92 | 345.84 | 78.23 | 179 | 14.00 |
| 2 | 226 | 0.00265 | 349.01 | 8.92 | 340.09 | 76.93 | 128 | 9.84 |
| 3 | 226 | 0.00221 | 335.27 | 8.92 | 326.35 | 73.82 | 77 | 5.66 |
| 4 | 226 | 0.00176 | 314.71 | 8.79 | 305.92 | 69.20 | 26 | 1.77 |
| 5 | 226 | 0.00131 | 262.35 | 7.86 | 254.49 | 57.56 | -26 | -1.47 |
| 6 | 226 | 0.00087 | 173.03 | 6.05 | 166.98 | 37.77 | -77 | -2.90 |
| 7 | 226 | 0.00042 | 83.70 | 3.34 | 80.36 | 18.18 | -128 | -2.32 |
| 8 | 226 | -0.00003 | -5.62 | 0.00 | -5.62 | -1.27 | -179 | 0.23 |
| Total | 410.41 | | 24.80 |
| xux | = | 401 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 401/450 = 0.321 |
| Puc | = | 0.321 x 20.00 x 300 x 450 |
| = | 865.69 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 865.69 + (410.41) | = | 1276.10 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 401/450 |
| = | 0.370 |
| Muc | = | 865.69 x (0.5 x 450 - 0.370 x 450) = 50.45 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 50.45 + (24.80) |
| = | 75.25 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 905 | 0.00292 | 353.22 | 8.92 | 344.30 | 311.52 | 104 | 32.40 |
| 2 | 905 | 0.00031 | 62.51 | 2.57 | 59.94 | 54.23 | -104 | -5.64 |
| Total | 365.75 | | 26.76 |
| xuy | = | 279 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 279/300 = 0.335 |
| Puc | = | 0.335 x 20.00 x 450 x 300 |
| = | 903.66 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 903.66 + (365.75) | = | 1269.41 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 279/300 |
| = | 0.387 |
| Muc | = | 903.66 x (0.5 x 300 - 0.387 x 300) = 30.70 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 30.70 + (26.76) |
| = | 57.46 kN-m |
| Pu/Puz | = | 0.718 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.718, | an | = | 1.864 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((26.20/75.25)1.864) + ((51.12/57.46)1.864) |
| = 0.944 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG9 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. (1810 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3375 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1620.97 kN |
| MomentX,(Mx) | = | 2.06 kN-m |
| MomentY,(My) | = | 34.62 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C17 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 35.256 kN.m |
| My_MinEccen | = | 32.419 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 35.256 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 34.619 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 2.06/1620.97 = 1 mm |
| Actual eccenY | = | My / P = 34.62/1620.97 = 21 mm |
| eccenXMin | = | (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(21,20) = 21 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(21 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00311 | 354.88 | 8.92 | 345.96 | 139.12 | 177 | 24.62 |
| 2 | 402 | 0.00263 | 348.56 | 8.92 | 339.64 | 136.58 | 118 | 16.12 |
| 3 | 402 | 0.00216 | 333.48 | 8.92 | 324.56 | 130.52 | 59 | 7.70 |
| 4 | 402 | 0.00168 | 309.75 | 8.69 | 301.06 | 121.06 | 0 | 0.00 |
| 5 | 402 | 0.00120 | 240.26 | 7.50 | 232.77 | 93.60 | -59 | -5.52 |
| 6 | 402 | 0.00072 | 144.76 | 5.29 | 139.47 | 56.08 | -118 | -6.62 |
| 7 | 402 | 0.00025 | 49.25 | 2.06 | 47.19 | 18.97 | -177 | -3.36 |
| Total | 695.93 | | 32.94 |
| xux | = | 432 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 432/450 = 0.346 |
| Puc | = | 0.346 x 20.00 x 300 x 450 |
| = | 934.03 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 934.03 + (695.93) | = | 1629.96 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 432/450 |
| = | 0.400 |
| Muc | = | 934.03 x (0.5 x 450 - 0.400 x 450) = 42.14 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 42.14 + (32.94) |
| = | 75.08 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1407 | 0.00291 | 353.16 | 8.92 | 344.24 | 484.49 | 102 | 49.42 |
| 2 | 1407 | 0.00060 | 119.67 | 4.54 | 115.13 | 162.03 | -102 | -16.53 |
| Total | 646.52 | | 32.89 |
| xuy | = | 305 mm Puc = C1.fck.B.D |
| ku | = | 1.016 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 1.083 |
| C1 | = | 0.446 x (1 - C3/6) = 0.366 |
| Puc | = | 0.366 x 20.00 x 450 x 300 |
| = | 986.88 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 986.88 + (646.52) | = | 1633.40 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.421 |
| Muc | = | 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 23.28 + (32.89) |
| = | 56.18 kN-m |
| Pu/Puz | = | 0.785 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.785, | an | = | 1.974 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((35.26/75.08)1.974) + ((34.62/56.18)1.974) |
| = 0.609 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG9 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG10 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 428.89 kN |
| MomentX,(Mx) | = | 58.03 kN-m |
| MomentY,(My) | = | 37.49 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C18 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 8.878 kN.m |
| My_MinEccen | = | 8.578 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 58.031 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 37.495 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 58.03/428.89 = 135 mm |
| Actual eccenY | = | My / P = 37.49/428.89 = 87 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(135,21) = 135 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(87,20) = 87 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(135 mm) > 0.05 x 300(15 mm)
and eccenY(87 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00274 | 351.36 | 8.92 | 342.44 | 77.46 | 179 | 13.87 |
| 2 | 226 | 0.00127 | 254.06 | 7.73 | 246.32 | 55.72 | 90 | 4.99 |
| 3 | 226 | -0.00020 | -40.50 | 0.00 | -40.50 | -9.16 | 0 | 0.00 |
| 4 | 226 | -0.00168 | -309.52 | 0.00 | -309.52 | -70.01 | -90 | 6.27 |
| 5 | 226 | -0.00315 | -355.20 | 0.00 | -355.20 | -80.34 | -179 | 14.38 |
| Total | -26.34 | | 39.50 |
| xux | = | 213 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 213/450 = 0.170 |
| Puc | = | 0.170 x 20.00 x 300 x 450 |
| = | 459.42 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 459.42 + (-26.34) | = | 433.08 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 213/450 |
| = | 0.197 |
| Muc | = | 459.42 x (0.5 x 450 - 0.197 x 450) = 62.72 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 62.72 + (39.50) |
| = | 102.22 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00233 | 339.71 | 8.92 | 330.79 | 187.05 | 104 | 19.45 |
| 2 | 565 | -0.00298 | -353.76 | 0.00 | -353.76 | -200.05 | -104 | 20.80 |
| Total | -12.99 | | 40.26 |
| xuy | = | 137 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 137/300 = 0.165 |
| Puc | = | 0.165 x 20.00 x 450 x 300 |
| = | 444.23 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 444.23 + (-12.99) | = | 431.24 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 137/300 |
| = | 0.190 |
| Muc | = | 444.23 x (0.5 x 300 - 0.190 x 300) = 41.30 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 41.30 + (40.26) |
| = | 81.56 kN-m |
| Pu/Puz | = | 0.275 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.275, | an | = | 1.126 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((58.03/102.22)1.126) + ((37.49/81.56)1.126) |
| = 0.946 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG10 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 843.23 kN |
| MomentX,(Mx) | = | 55.96 kN-m |
| MomentY,(My) | = | 33.28 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C18 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 17.455 kN.m |
| My_MinEccen | = | 16.865 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 55.962 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 33.275 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 55.96/843.23 = 66 mm |
| Actual eccenY | = | My / P = 33.28/843.23 = 39 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(66,21) = 66 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(39,20) = 39 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(66 mm) > 0.05 x 300(15 mm)
and eccenY(39 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00300 | 353.87 | 8.92 | 344.95 | 78.03 | 179 | 13.97 |
| 2 | 226 | 0.00202 | 328.39 | 8.92 | 319.47 | 72.26 | 90 | 6.47 |
| 3 | 226 | 0.00104 | 207.69 | 6.86 | 200.83 | 45.43 | 0 | 0.00 |
| 4 | 226 | 0.00006 | 11.86 | 0.52 | 11.34 | 2.57 | -90 | -0.23 |
| 5 | 226 | -0.00092 | -183.97 | 0.00 | -183.97 | -41.61 | -179 | 7.45 |
| Total | 156.67 | | 27.65 |
| xux | = | 320 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 320/450 = 0.256 |
| Puc | = | 0.256 x 20.00 x 300 x 450 |
| = | 691.03 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 691.03 + (156.67) | = | 847.70 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 320/450 |
| = | 0.296 |
| Muc | = | 691.03 x (0.5 x 450 - 0.296 x 450) = 63.51 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 63.51 + (27.65) |
| = | 91.17 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00277 | 351.88 | 8.92 | 342.96 | 193.94 | 104 | 20.17 |
| 2 | 565 | -0.00054 | -107.04 | 0.00 | -107.04 | -60.53 | -104 | 6.29 |
| Total | 133.41 | | 26.46 |
| xuy | = | 220 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 220/300 = 0.264 |
| Puc | = | 0.264 x 20.00 x 450 x 300 |
| = | 713.81 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 713.81 + (133.41) | = | 847.23 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 220/300 |
| = | 0.306 |
| Muc | = | 713.81 x (0.5 x 300 - 0.306 x 300) = 41.65 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 41.65 + (26.46) |
| = | 68.12 kN-m |
| Pu/Puz | = | 0.542 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.542, | an | = | 1.569 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((55.96/91.17)1.569) + ((33.28/68.12)1.569) |
| = 0.790 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG10 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1264.49 kN |
| MomentX,(Mx) | = | 49.63 kN-m |
| MomentY,(My) | = | 33.41 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C18 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 26.175 kN.m |
| My_MinEccen | = | 25.290 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 49.629 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 33.410 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 49.63/1264.49 = 39 mm |
| Actual eccenY | = | My / P = 33.41/1264.49 = 26 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(39,21) = 39 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(26,20) = 26 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(39 mm) > 0.05 x 300(15 mm)
and eccenY(26 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 8 nos. (1608 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00310 | 354.76 | 8.92 | 345.84 | 139.07 | 177 | 24.62 |
| 2 | 402 | 0.00211 | 331.83 | 8.92 | 322.91 | 129.85 | 59 | 7.66 |
| 3 | 402 | 0.00112 | 224.81 | 7.21 | 217.60 | 87.50 | -59 | -5.16 |
| 4 | 402 | 0.00014 | 27.37 | 1.18 | 26.19 | 10.53 | -177 | -1.86 |
| Total | 366.95 | | 25.25 |
| xux | = | 418 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 418/450 = 0.335 |
| Puc | = | 0.335 x 20.00 x 300 x 450 |
| = | 903.66 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 903.66 + (366.95) | = | 1270.61 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 418/450 |
| = | 0.387 |
| Muc | = | 903.66 x (0.5 x 450 - 0.387 x 450) = 46.05 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 46.05 + (25.25) |
| = | 71.30 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 804 | 0.00291 | 353.13 | 8.92 | 344.21 | 276.83 | 102 | 28.24 |
| 2 | 804 | 0.00042 | 83.08 | 3.32 | 79.76 | 64.15 | -102 | -6.54 |
| Total | 340.98 | | 21.69 |
| xuy | = | 286 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 286/300 = 0.343 |
| Puc | = | 0.343 x 20.00 x 450 x 300 |
| = | 926.44 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 926.44 + (340.98) | = | 1267.42 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 286/300 |
| = | 0.397 |
| Muc | = | 926.44 x (0.5 x 300 - 0.397 x 300) = 28.77 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 28.77 + (21.69) |
| = | 50.46 kN-m |
| Pu/Puz | = | 0.743 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.743, | an | = | 1.906 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((49.63/71.30)1.906) + ((33.41/50.46)1.906) |
| = 0.957 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG10 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 8 nos. (1608 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1732.79 kN |
| MomentX,(Mx) | = | 38.87 kN-m |
| MomentY,(My) | = | 34.58 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C18 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 35.869 kN.m |
| My_MinEccen | = | 34.656 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 38.873 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 34.656 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 38.87/1732.79 = 22 mm |
| Actual eccenY | = | My / P = 34.58/1732.79 = 20 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(22,21) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(20,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00311 | 354.90 | 8.92 | 345.98 | 139.13 | 177 | 24.63 |
| 2 | 402 | 0.00271 | 350.48 | 8.92 | 341.56 | 137.35 | 126 | 17.36 |
| 3 | 402 | 0.00230 | 338.85 | 8.92 | 329.93 | 132.67 | 76 | 10.06 |
| 4 | 402 | 0.00190 | 323.34 | 8.90 | 314.44 | 126.44 | 25 | 3.20 |
| 5 | 402 | 0.00149 | 293.49 | 8.34 | 285.15 | 114.66 | -25 | -2.90 |
| 6 | 402 | 0.00108 | 216.90 | 7.05 | 209.85 | 84.39 | -76 | -6.40 |
| 7 | 402 | 0.00068 | 135.70 | 5.03 | 130.67 | 52.55 | -126 | -6.64 |
| 8 | 402 | 0.00027 | 54.49 | 2.26 | 52.23 | 21.00 | -177 | -3.72 |
| Total | 808.20 | | 35.59 |
| xux | = | 436 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 436/450 = 0.349 |
| Puc | = | 0.349 x 20.00 x 300 x 450 |
| = | 941.63 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 941.63 + (808.20) | = | 1749.82 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 436/450 |
| = | 0.403 |
| Muc | = | 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 41.10 + (35.59) |
| = | 76.69 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1608 | 0.00289 | 352.95 | 8.92 | 344.03 | 553.37 | 102 | 56.44 |
| 2 | 1608 | 0.00063 | 126.93 | 4.76 | 122.17 | 196.51 | -102 | -20.04 |
| Total | 749.88 | | 36.40 |
| xuy | = | 309 mm Puc = C1.fck.B.D |
| ku | = | 1.031 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 1.027 |
| C1 | = | 0.446 x (1 - C3/6) = 0.370 |
| Puc | = | 0.370 x 20.00 x 450 x 300 |
| = | 998.00 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 998.00 + (749.88) | = | 1747.88 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.426 |
| Muc | = | 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 22.09 + (36.40) |
| = | 58.49 kN-m |
| Pu/Puz | = | 0.792 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.792, | an | = | 1.987 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((38.87/76.69)1.987) + ((34.66/58.49)1.987) |
| = 0.613 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG10 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3375 mm |
From analysis results,loads on column
| Axial load,(P) | = | 2203.72 kN |
| MomentX,(Mx) | = | 20.93 kN-m |
| MomentY,(My) | = | 22.73 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C18 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 47.931 kN.m |
| My_MinEccen | = | 44.074 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 47.931 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 44.074 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 20.93/2203.72 = 9 mm |
| Actual eccenY | = | My / P = 22.73/2203.72 = 10 mm |
| eccenXMin | = | (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(9,22) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(10,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 628 | 0.00310 | 354.76 | 8.92 | 345.84 | 217.30 | 175 | 38.03 |
| 2 | 628 | 0.00270 | 350.18 | 8.92 | 341.26 | 214.42 | 125 | 26.80 |
| 3 | 628 | 0.00230 | 338.60 | 8.92 | 329.68 | 207.14 | 75 | 15.54 |
| 4 | 628 | 0.00189 | 323.19 | 8.90 | 314.30 | 197.48 | 25 | 4.94 |
| 5 | 628 | 0.00149 | 293.71 | 8.35 | 285.36 | 179.30 | -25 | -4.48 |
| 6 | 628 | 0.00109 | 218.28 | 7.08 | 211.20 | 132.70 | -75 | -9.95 |
| 7 | 628 | 0.00069 | 137.99 | 5.09 | 132.90 | 83.50 | -125 | -10.44 |
| 8 | 628 | 0.00029 | 57.71 | 2.39 | 55.32 | 34.76 | -175 | -6.08 |
| Total | 1266.60 | | 54.35 |
| xux | = | 436 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 436/450 = 0.349 |
| Puc | = | 0.349 x 20.00 x 300 x 450 |
| = | 941.63 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 941.63 + (1266.60) | = | 2208.23 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 436/450 |
| = | 0.403 |
| Muc | = | 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 41.10 + (54.35) |
| = | 95.45 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 2513 | 0.00285 | 352.56 | 8.92 | 343.64 | 863.67 | 100 | 86.37 |
| 2 | 2513 | 0.00069 | 138.15 | 5.10 | 133.05 | 334.39 | -100 | -33.44 |
| Total | 1198.06 | | 52.93 |
| xuy | = | 314 mm Puc = C1.fck.B.D |
| ku | = | 1.047 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 0.976 |
| C1 | = | 0.446 x (1 - C3/6) = 0.373 |
| Puc | = | 0.373 x 20.00 x 450 x 300 |
| = | 1008.29 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 1008.29 + (1198.06) | = | 2206.35 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.431 |
| Muc | = | 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 20.99 + (52.93) |
| = | 73.92 kN-m |
| Pu/Puz | = | 0.806 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.806, | an | = | 2.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((47.93/95.45)2.000) + ((44.07/73.92)2.000) |
| = 0.608 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 20 = 320 mm
iii) 300 mm
Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties 8 @ 300 mm c/c
SUMMARY :
CG10 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG11 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 558.65 kN |
| MomentX,(Mx) | = | 34.86 kN-m |
| MomentY,(My) | = | 56.09 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C19 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 11.564 kN.m |
| My_MinEccen | = | 11.173 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 34.862 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 56.087 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 34.86/558.65 = 62 mm |
| Actual eccenY | = | My / P = 56.09/558.65 = 100 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(62,21) = 62 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(100,20) = 100 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(62 mm) > 0.05 x 300(15 mm)
and eccenY(100 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00284 | 352.51 | 8.92 | 343.59 | 77.72 | 179 | 13.91 |
| 2 | 226 | 0.00156 | 299.98 | 8.49 | 291.49 | 65.93 | 90 | 5.90 |
| 3 | 226 | 0.00028 | 55.40 | 2.30 | 53.10 | 12.01 | 0 | 0.00 |
| 4 | 226 | -0.00101 | -201.01 | 0.00 | -201.01 | -45.47 | -90 | 4.07 |
| 5 | 226 | -0.00229 | -338.29 | 0.00 | -338.29 | -76.52 | -179 | 13.70 |
| Total | 33.67 | | 37.58 |
| xux | = | 244 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 244/450 = 0.195 |
| Puc | = | 0.195 x 20.00 x 300 x 450 |
| = | 527.77 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 527.77 + (33.67) | = | 561.44 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 244/450 |
| = | 0.226 |
| Muc | = | 527.77 x (0.5 x 450 - 0.226 x 450) = 65.10 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 65.10 + (37.58) |
| = | 102.68 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00255 | 346.30 | 8.92 | 337.38 | 190.78 | 104 | 19.84 |
| 2 | 565 | -0.00177 | -315.32 | 0.00 | -315.32 | -178.31 | -104 | 18.54 |
| Total | 12.47 | | 38.39 |
| xuy | = | 169 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 169/300 = 0.203 |
| Puc | = | 0.203 x 20.00 x 450 x 300 |
| = | 546.75 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 546.75 + (12.47) | = | 559.22 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 169/300 |
| = | 0.234 |
| Muc | = | 546.75 x (0.5 x 300 - 0.234 x 300) = 43.63 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 43.63 + (38.39) |
| = | 82.02 kN-m |
| Pu/Puz | = | 0.359 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.359, | an | = | 1.265 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((34.86/102.68)1.265) + ((56.09/82.02)1.265) |
| = 0.873 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG11 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1113.06 kN |
| MomentX,(Mx) | = | 36.77 kN-m |
| MomentY,(My) | = | 51.38 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C19 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 23.040 kN.m |
| My_MinEccen | = | 22.261 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 36.770 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 51.377 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 36.77/1113.06 = 33 mm |
| Actual eccenY | = | My / P = 51.38/1113.06 = 46 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(33,21) = 33 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(46,20) = 46 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(33 mm) > 0.05 x 300(15 mm)
and eccenY(46 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00306 | 354.44 | 8.92 | 345.52 | 138.94 | 177 | 24.59 |
| 2 | 226 | 0.00224 | 336.71 | 8.92 | 327.79 | 74.14 | 88 | 6.49 |
| 3 | 226 | 0.00144 | 289.17 | 8.23 | 280.94 | 63.55 | 0 | 0.00 |
| 4 | 226 | 0.00065 | 129.15 | 4.83 | 124.32 | 28.12 | -88 | -2.46 |
| 5 | 402 | -0.00017 | -34.34 | 0.00 | -34.34 | -13.81 | -177 | 2.44 |
| Total | 290.95 | | 31.06 |
| xux | = | 383 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 383/450 = 0.307 |
| Puc | = | 0.307 x 20.00 x 300 x 450 |
| = | 827.72 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 827.72 + (290.95) | = | 1118.66 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 383/450 |
| = | 0.354 |
| Muc | = | 827.72 x (0.5 x 450 - 0.354 x 450) = 54.29 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 54.29 + (31.06) |
| = | 85.35 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 741 | 0.00285 | 352.62 | 8.92 | 343.70 | 254.83 | 102 | 25.99 |
| 2 | 741 | 0.00011 | 21.95 | 0.95 | 20.99 | 15.57 | -102 | -1.59 |
| Total | 270.39 | | 24.40 |
| xuy | = | 260 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 260/300 = 0.312 |
| Puc | = | 0.312 x 20.00 x 450 x 300 |
| = | 842.91 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 842.91 + (270.39) | = | 1113.30 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 260/300 |
| = | 0.361 |
| Muc | = | 842.91 x (0.5 x 300 - 0.361 x 300) = 35.21 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 35.21 + (24.40) |
| = | 59.62 kN-m |
| Pu/Puz | = | 0.669 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.669, | an | = | 1.782 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((36.77/85.35)1.782) + ((51.38/59.62)1.782) |
| = 0.990 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG11 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1685.34 kN |
| MomentX,(Mx) | = | 22.68 kN-m |
| MomentY,(My) | = | 54.55 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C19 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 34.887 kN.m |
| My_MinEccen | = | 33.707 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 34.887 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 54.547 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 22.68/1685.34 = 13 mm |
| Actual eccenY | = | My / P = 54.55/1685.34 = 32 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(13,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(32,20) = 32 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(32 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00310 | 354.79 | 8.92 | 345.87 | 139.08 | 177 | 24.62 |
| 2 | 402 | 0.00268 | 349.80 | 8.92 | 340.88 | 137.08 | 126 | 17.33 |
| 3 | 402 | 0.00226 | 337.39 | 8.92 | 328.47 | 132.08 | 76 | 10.02 |
| 4 | 402 | 0.00184 | 320.00 | 8.87 | 311.14 | 125.12 | 25 | 3.16 |
| 5 | 402 | 0.00142 | 284.71 | 8.18 | 276.53 | 111.20 | -25 | -2.81 |
| 6 | 402 | 0.00100 | 200.80 | 6.71 | 194.09 | 78.05 | -76 | -5.92 |
| 7 | 402 | 0.00058 | 116.89 | 4.45 | 112.44 | 45.21 | -126 | -5.72 |
| 8 | 402 | 0.00016 | 32.98 | 1.41 | 31.57 | 12.69 | -177 | -2.25 |
| Total | 780.52 | | 38.44 |
| xux | = | 422 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 422/450 = 0.338 |
| Puc | = | 0.338 x 20.00 x 300 x 450 |
| = | 911.25 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 911.25 + (780.52) | = | 1691.77 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 422/450 |
| = | 0.390 |
| Muc | = | 911.25 x (0.5 x 450 - 0.390 x 450) = 45.11 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 45.11 + (38.44) |
| = | 83.54 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1608 | 0.00294 | 353.38 | 8.92 | 344.45 | 554.05 | 102 | 56.51 |
| 2 | 1608 | 0.00056 | 112.00 | 4.30 | 107.70 | 173.24 | -102 | -17.67 |
| Total | 727.30 | | 38.84 |
| xuy | = | 300 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 300/300 = 0.360 |
| Puc | = | 0.360 x 20.00 x 450 x 300 |
| = | 972.00 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 972.00 + (727.30) | = | 1699.30 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 300/300 |
| = | 0.416 |
| Muc | = | 972.00 x (0.5 x 300 - 0.416 x 300) = 24.49 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 24.49 + (38.84) |
| = | 63.34 kN-m |
| Pu/Puz | = | 0.770 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.770, | an | = | 1.951 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((34.89/83.54)1.951) + ((54.55/63.34)1.951) |
| = 0.929 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG11 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 2226.13 kN |
| MomentX,(Mx) | = | 10.01 kN-m |
| MomentY,(My) | = | 59.82 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C19 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 46.081 kN.m |
| My_MinEccen | = | 44.523 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 46.081 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 59.817 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 10.01/2226.13 = 4 mm |
| Actual eccenY | = | My / P = 59.82/2226.13 = 27 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(4,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(27,20) = 27 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(27 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 628 | 0.00310 | 354.82 | 8.92 | 345.90 | 217.33 | 175 | 38.03 |
| 2 | 628 | 0.00271 | 350.51 | 8.92 | 341.59 | 214.63 | 125 | 26.83 |
| 3 | 628 | 0.00231 | 339.30 | 8.92 | 330.38 | 207.59 | 75 | 15.57 |
| 4 | 628 | 0.00192 | 324.78 | 8.91 | 315.88 | 198.47 | 25 | 4.96 |
| 5 | 628 | 0.00152 | 296.72 | 8.42 | 288.31 | 181.15 | -25 | -4.53 |
| 6 | 628 | 0.00113 | 225.93 | 7.23 | 218.70 | 137.41 | -75 | -10.31 |
| 7 | 628 | 0.00073 | 146.91 | 5.35 | 141.56 | 88.95 | -125 | -11.12 |
| 8 | 628 | 0.00034 | 67.90 | 2.77 | 65.13 | 40.92 | -175 | -7.16 |
| Total | 1286.45 | | 52.28 |
| xux | = | 443 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 443/450 = 0.354 |
| Puc | = | 0.354 x 20.00 x 300 x 450 |
| = | 956.81 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 956.81 + (1286.45) | = | 2243.26 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 443/450 |
| = | 0.410 |
| Muc | = | 956.81 x (0.5 x 450 - 0.410 x 450) = 38.97 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 38.97 + (52.28) |
| = | 91.24 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 2513 | 0.00283 | 352.38 | 8.92 | 343.46 | 863.21 | 100 | 86.32 |
| 2 | 2513 | 0.00072 | 144.60 | 5.28 | 139.32 | 350.14 | -100 | -35.01 |
| Total | 1213.35 | | 51.31 |
| xuy | = | 319 mm Puc = C1.fck.B.D |
| ku | = | 1.063 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 0.929 |
| C1 | = | 0.446 x (1 - C3/6) = 0.377 |
| Puc | = | 0.377 x 20.00 x 450 x 300 |
| = | 1017.83 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 1017.83 + (1213.35) | = | 2231.18 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.435 |
| Muc | = | 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 19.97 + (51.31) |
| = | 71.28 kN-m |
| Pu/Puz | = | 0.814 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.814, | an | = | 2.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((46.08/91.24)2.000) + ((59.82/71.28)2.000) |
| = 0.959 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 20 = 320 mm
iii) 300 mm
Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties 8 @ 300 mm c/c
SUMMARY :
CG11 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3375 mm |
From analysis results,loads on column
| Axial load,(P) | = | 2853.12 kN |
| MomentX,(Mx) | = | 14.31 kN-m |
| MomentY,(My) | = | 41.16 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C19 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 62.055 kN.m |
| My_MinEccen | = | 57.062 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 62.055 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 57.062 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 14.31/2853.12 = 5 mm |
| Actual eccenY | = | My / P = 41.16/2853.12 = 14 mm |
| eccenXMin | = | (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(5,22) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(14,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 10 nos. (8042 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1608 | 0.00304 | 354.24 | 8.92 | 345.32 | 555.45 | 169 | 93.87 |
| 2 | 1608 | 0.00234 | 340.38 | 8.92 | 331.46 | 533.14 | 85 | 45.05 |
| 3 | 1608 | 0.00165 | 307.87 | 8.64 | 299.23 | 481.30 | 0 | 0.00 |
| 4 | 1608 | 0.00095 | 190.70 | 6.48 | 184.23 | 296.33 | -85 | -25.04 |
| 5 | 1608 | 0.00026 | 51.65 | 2.16 | 49.50 | 79.62 | -169 | -13.46 |
| Total | 1945.85 | | 100.43 |
| xux | = | 425 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 425/450 = 0.340 |
| Puc | = | 0.340 x 20.00 x 300 x 450 |
| = | 918.84 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 918.84 + (1945.85) | = | 2864.69 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 425/450 |
| = | 0.393 |
| Muc | = | 918.84 x (0.5 x 450 - 0.393 x 450) = 44.14 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 44.14 + (100.43) |
| = | 144.57 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 4021 | 0.00285 | 352.56 | 8.92 | 343.64 | 1381.85 | 94 | 129.89 |
| 2 | 4021 | 0.00065 | 130.67 | 4.88 | 125.79 | 505.83 | -94 | -47.55 |
| Total | 1887.69 | | 82.35 |
| xuy | = | 300 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 300/300 = 0.360 |
| Puc | = | 0.360 x 20.00 x 450 x 300 |
| = | 972.00 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 972.00 + (1887.69) | = | 2859.69 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 300/300 |
| = | 0.416 |
| Muc | = | 972.00 x (0.5 x 300 - 0.416 x 300) = 24.49 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 24.49 + (82.35) |
| = | 106.84 kN-m |
| Pu/Puz | = | 0.783 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.783, | an | = | 1.971 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((62.06/144.57)1.971) + ((57.06/106.84)1.971) |
| = 0.479 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 32 = 512 mm
iii) 300 mm
Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties 8 @ 300 mm c/c
SUMMARY :
CG11 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 10 nos. (8042 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG12 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 386.62 kN |
| MomentX,(Mx) | = | 7.65 kN-m |
| MomentY,(My) | = | 85.25 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C20 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 8.003 kN.m |
| My_MinEccen | = | 7.732 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 8.003 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 85.251 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 7.65/386.62 = 20 mm |
| Actual eccenY | = | My / P = 85.25/386.62 = 221 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(20,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(221,20) = 221 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(221 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00267 | 349.46 | 8.92 | 340.54 | 136.94 | 177 | 24.24 |
| 2 | 226 | 0.00112 | 223.86 | 7.19 | 216.67 | 49.01 | 88 | 4.29 |
| 3 | 226 | -0.00040 | -79.13 | 0.00 | -79.13 | -17.90 | 0 | 0.00 |
| 4 | 226 | -0.00191 | -324.22 | 0.00 | -324.22 | -73.34 | -88 | 6.42 |
| 5 | 402 | -0.00346 | -357.93 | 0.00 | -357.93 | -143.93 | -177 | 25.48 |
| Total | -49.22 | | 60.42 |
| xux | = | 202 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 202/450 = 0.162 |
| Puc | = | 0.162 x 20.00 x 300 x 450 |
| = | 436.64 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 436.64 + (-49.22) | = | 387.42 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 202/450 |
| = | 0.187 |
| Muc | = | 436.64 x (0.5 x 450 - 0.187 x 450) = 61.53 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 61.53 + (60.42) |
| = | 121.94 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 741 | 0.00218 | 334.53 | 8.92 | 325.61 | 241.41 | 102 | 24.62 |
| 2 | 741 | -0.00340 | -357.44 | 0.00 | -357.44 | -265.01 | -102 | 27.03 |
| Total | -23.60 | | 51.66 |
| xuy | = | 128 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 128/300 = 0.153 |
| Puc | = | 0.153 x 20.00 x 450 x 300 |
| = | 413.86 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 413.86 + (-23.60) | = | 390.25 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 128/300 |
| = | 0.177 |
| Muc | = | 413.86 x (0.5 x 300 - 0.177 x 300) = 40.09 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 40.09 + (51.66) |
| = | 91.74 kN-m |
| Pu/Puz | = | 0.232 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.232, | an | = | 1.054 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((8.00/121.94)1.054) + ((85.25/91.74)1.054) |
| = 0.982 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG12 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 781.33 kN |
| MomentX,(Mx) | = | 7.26 kN-m |
| MomentY,(My) | = | 77.98 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C20 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 16.174 kN.m |
| My_MinEccen | = | 15.627 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 16.174 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 77.977 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 7.26/781.33 = 9 mm |
| Actual eccenY | = | My / P = 77.98/781.33 = 100 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(9,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(100,20) = 100 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(100 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 14 nos. (1583 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00294 | 353.42 | 8.92 | 344.50 | 77.92 | 179 | 13.95 |
| 2 | 226 | 0.00222 | 336.00 | 8.92 | 327.08 | 73.98 | 119 | 8.83 |
| 3 | 226 | 0.00150 | 294.85 | 8.37 | 286.47 | 64.80 | 60 | 3.87 |
| 4 | 226 | 0.00078 | 156.97 | 5.63 | 151.34 | 34.23 | -0 | -0.00 |
| 5 | 226 | 0.00006 | 12.97 | 0.57 | 12.40 | 2.80 | -60 | -0.17 |
| 6 | 226 | -0.00066 | -131.04 | 0.00 | -131.04 | -29.64 | -119 | 3.54 |
| 7 | 226 | -0.00138 | -275.04 | 0.00 | -275.04 | -62.21 | -179 | 11.14 |
| Total | 161.89 | | 41.15 |
| xux | = | 290 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 290/450 = 0.232 |
| Puc | = | 0.232 x 20.00 x 300 x 450 |
| = | 626.48 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 626.48 + (161.89) | = | 788.37 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 290/450 |
| = | 0.268 |
| Muc | = | 626.48 x (0.5 x 450 - 0.268 x 450) = 65.37 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 65.37 + (41.15) |
| = | 106.52 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 792 | 0.00271 | 350.41 | 8.92 | 341.49 | 270.35 | 104 | 28.12 |
| 2 | 792 | -0.00089 | -177.01 | 0.00 | -177.01 | -140.14 | -104 | 14.57 |
| Total | 130.21 | | 42.69 |
| xuy | = | 203 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 203/300 = 0.243 |
| Puc | = | 0.243 x 20.00 x 450 x 300 |
| = | 656.86 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 656.86 + (130.21) | = | 787.07 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 203/300 |
| = | 0.281 |
| Muc | = | 656.86 x (0.5 x 300 - 0.281 x 300) = 43.13 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 43.13 + (42.69) |
| = | 85.82 kN-m |
| Pu/Puz | = | 0.461 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.461, | an | = | 1.436 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((16.17/106.52)1.436) + ((77.98/85.82)1.436) |
| = 0.938 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG12 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 14 nos. (1583 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1185.41 kN |
| MomentX,(Mx) | = | 5.82 kN-m |
| MomentY,(My) | = | 77.19 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C20 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 24.538 kN.m |
| My_MinEccen | = | 23.708 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 24.538 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 77.191 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 5.82/1185.41 = 5 mm |
| Actual eccenY | = | My / P = 77.19/1185.41 = 65 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(5,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(65,20) = 65 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(65 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00300 | 353.90 | 8.92 | 344.98 | 138.72 | 177 | 24.55 |
| 2 | 402 | 0.00238 | 341.87 | 8.92 | 332.95 | 133.89 | 118 | 15.80 |
| 3 | 402 | 0.00177 | 315.41 | 8.80 | 306.61 | 123.29 | 59 | 7.27 |
| 4 | 402 | 0.00115 | 230.89 | 7.33 | 223.56 | 89.90 | 0 | 0.00 |
| 5 | 402 | 0.00054 | 107.88 | 4.16 | 103.72 | 41.71 | -59 | -2.46 |
| 6 | 402 | -0.00008 | -15.13 | 0.00 | -15.13 | -6.08 | -118 | 0.72 |
| 7 | 402 | -0.00069 | -138.14 | 0.00 | -138.14 | -55.55 | -177 | 9.83 |
| Total | 465.87 | | 55.72 |
| xux | = | 336 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 336/450 = 0.269 |
| Puc | = | 0.269 x 20.00 x 300 x 450 |
| = | 725.20 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 725.20 + (465.87) | = | 1191.08 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 336/450 |
| = | 0.310 |
| Muc | = | 725.20 x (0.5 x 450 - 0.310 x 450) = 61.88 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 61.88 + (55.72) |
| = | 117.60 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1407 | 0.00279 | 352.07 | 8.92 | 343.15 | 482.95 | 102 | 49.26 |
| 2 | 1407 | -0.00023 | -45.19 | 0.00 | -45.19 | -63.60 | -102 | 6.49 |
| Total | 419.35 | | 55.75 |
| xuy | = | 237 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 237/300 = 0.284 |
| Puc | = | 0.284 x 20.00 x 450 x 300 |
| = | 766.97 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 766.97 + (419.35) | = | 1186.32 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 237/300 |
| = | 0.328 |
| Muc | = | 766.97 x (0.5 x 300 - 0.328 x 300) = 39.52 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 39.52 + (55.75) |
| = | 95.27 kN-m |
| Pu/Puz | = | 0.574 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.574, | an | = | 1.623 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((24.54/117.60)1.623) + ((77.19/95.27)1.623) |
| = 0.789 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG12 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1604.80 kN |
| MomentX,(Mx) | = | 3.94 kN-m |
| MomentY,(My) | = | 76.38 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C20 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 33.219 kN.m |
| My_MinEccen | = | 32.096 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 33.219 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 76.379 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 3.94/1604.80 = 2 mm |
| Actual eccenY | = | My / P = 76.38/1604.80 = 48 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(48,20) = 48 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(48 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 12 nos. (3770 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 628 | 0.00304 | 354.26 | 8.92 | 345.34 | 216.98 | 175 | 37.97 |
| 2 | 628 | 0.00240 | 342.39 | 8.92 | 333.47 | 209.53 | 105 | 22.00 |
| 3 | 628 | 0.00176 | 314.60 | 8.79 | 305.82 | 192.15 | 35 | 6.73 |
| 4 | 628 | 0.00111 | 222.87 | 7.17 | 215.70 | 135.53 | -35 | -4.74 |
| 5 | 628 | 0.00047 | 94.41 | 3.71 | 90.69 | 56.99 | -105 | -5.98 |
| 6 | 628 | -0.00017 | -34.05 | 0.00 | -34.05 | -21.39 | -175 | 3.74 |
| Total | 789.78 | | 59.71 |
| xux | = | 381 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 381/450 = 0.305 |
| Puc | = | 0.305 x 20.00 x 300 x 450 |
| = | 823.92 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 823.92 + (789.78) | = | 1613.70 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 381/450 |
| = | 0.353 |
| Muc | = | 823.92 x (0.5 x 450 - 0.353 x 450) = 54.64 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 54.64 + (59.71) |
| = | 114.36 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1885 | 0.00285 | 352.59 | 8.92 | 343.67 | 647.81 | 100 | 64.78 |
| 2 | 1885 | 0.00025 | 50.72 | 2.12 | 48.61 | 91.62 | -100 | -9.16 |
| Total | 739.43 | | 55.62 |
| xuy | = | 270 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 270/300 = 0.323 |
| Puc | = | 0.323 x 20.00 x 450 x 300 |
| = | 873.28 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 873.28 + (739.43) | = | 1612.71 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 270/300 |
| = | 0.374 |
| Muc | = | 873.28 x (0.5 x 300 - 0.374 x 300) = 33.08 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 33.08 + (55.62) |
| = | 88.69 kN-m |
| Pu/Puz | = | 0.682 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.682, | an | = | 1.803 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((33.22/114.36)1.803) + ((76.38/88.69)1.803) |
| = 0.871 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 20 = 320 mm
iii) 300 mm
Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties 8 @ 300 mm c/c
SUMMARY :
CG12 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 12 nos. (3770 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3375 mm |
From analysis results,loads on column
| Axial load,(P) | = | 2044.35 kN |
| MomentX,(Mx) | = | 1.14 kN-m |
| MomentY,(My) | = | 51.67 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C20 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 44.465 kN.m |
| My_MinEccen | = | 40.887 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 44.465 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 51.666 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 1.14/2044.35 = 1 mm |
| Actual eccenY | = | My / P = 51.67/2044.35 = 25 mm |
| eccenXMin | = | (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(25,20) = 25 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 628 | 0.00310 | 354.79 | 8.92 | 345.87 | 217.32 | 175 | 38.03 |
| 2 | 628 | 0.00264 | 348.64 | 8.92 | 339.72 | 213.45 | 117 | 24.90 |
| 3 | 628 | 0.00217 | 334.08 | 8.92 | 325.16 | 204.30 | 58 | 11.92 |
| 4 | 628 | 0.00171 | 311.57 | 8.73 | 302.84 | 190.28 | 0 | 0.00 |
| 5 | 628 | 0.00124 | 248.68 | 7.64 | 241.04 | 151.45 | -58 | -8.83 |
| 6 | 628 | 0.00078 | 155.76 | 5.59 | 150.17 | 94.35 | -117 | -11.01 |
| 7 | 628 | 0.00031 | 62.84 | 2.58 | 60.26 | 37.86 | -175 | -6.63 |
| Total | 1109.02 | | 48.38 |
| xux | = | 439 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 439/450 = 0.352 |
| Puc | = | 0.352 x 20.00 x 300 x 450 |
| = | 949.22 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 949.22 + (1109.02) | = | 2058.24 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 439/450 |
| = | 0.406 |
| Muc | = | 949.22 x (0.5 x 450 - 0.406 x 450) = 40.05 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 40.05 + (48.38) |
| = | 88.43 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 2199 | 0.00285 | 352.56 | 8.92 | 343.64 | 755.71 | 100 | 75.57 |
| 2 | 2199 | 0.00069 | 138.15 | 5.10 | 133.05 | 292.59 | -100 | -29.26 |
| Total | 1048.30 | | 46.31 |
| xuy | = | 314 mm Puc = C1.fck.B.D |
| ku | = | 1.047 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 0.976 |
| C1 | = | 0.446 x (1 - C3/6) = 0.373 |
| Puc | = | 0.373 x 20.00 x 450 x 300 |
| = | 1008.29 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 1008.29 + (1048.30) | = | 2056.59 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.431 |
| Muc | = | 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 20.99 + (46.31) |
| = | 67.30 kN-m |
| Pu/Puz | = | 0.803 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.803, | an | = | 2.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((44.46/88.43)2.000) + ((51.67/67.30)2.000) |
| = 0.842 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 20 = 320 mm
iii) 300 mm
Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties 8 @ 300 mm c/c
SUMMARY :
CG12 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG13 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 283.57 kN |
| MomentX,(Mx) | = | 18.07 kN-m |
| MomentY,(My) | = | 44.39 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C21 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 5.870 kN.m |
| My_MinEccen | = | 5.671 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 18.070 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 44.386 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 18.07/283.57 = 64 mm |
| Actual eccenY | = | My / P = 44.39/283.57 = 157 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(64,21) = 64 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(157,20) = 157 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(64 mm) > 0.05 x 300(15 mm)
and eccenY(157 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00258 | 347.28 | 8.92 | 338.36 | 76.53 | 179 | 13.70 |
| 2 | 226 | 0.00080 | 160.41 | 5.72 | 154.69 | 34.99 | 90 | 3.13 |
| 3 | 226 | -0.00098 | -196.00 | 0.00 | -196.00 | -44.33 | 0 | 0.00 |
| 4 | 226 | -0.00276 | -351.82 | 0.00 | -351.82 | -79.58 | -90 | 7.12 |
| 5 | 226 | -0.00454 | -360.90 | 0.00 | -360.90 | -81.63 | -179 | 14.61 |
| Total | -94.02 | | 38.57 |
| xux | = | 176 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 176/450 = 0.141 |
| Puc | = | 0.141 x 20.00 x 300 x 450 |
| = | 379.69 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 379.69 + (-94.02) | = | 285.66 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 176/450 |
| = | 0.163 |
| Muc | = | 379.69 x (0.5 x 450 - 0.163 x 450) = 57.67 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 57.67 + (38.57) |
| = | 96.23 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00183 | 319.48 | 8.86 | 310.62 | 175.65 | 104 | 18.27 |
| 2 | 565 | -0.00570 | -360.90 | 0.00 | -360.90 | -204.08 | -104 | 21.22 |
| Total | -28.43 | | 39.49 |
| xuy | = | 97 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 97/300 = 0.116 |
| Puc | = | 0.116 x 20.00 x 450 x 300 |
| = | 313.24 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 313.24 + (-28.43) | = | 284.81 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 97/300 |
| = | 0.134 |
| Muc | = | 313.24 x (0.5 x 300 - 0.134 x 300) = 34.39 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 34.39 + (39.49) |
| = | 73.88 kN-m |
| Pu/Puz | = | 0.182 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.182, | an | = | 1.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((18.07/96.23)1.000) + ((44.39/73.88)1.000) |
| = 0.789 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG13 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 586.34 kN |
| MomentX,(Mx) | = | 16.25 kN-m |
| MomentY,(My) | = | 40.17 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C21 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 12.137 kN.m |
| My_MinEccen | = | 11.727 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 16.248 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 40.171 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 16.25/586.34 = 28 mm |
| Actual eccenY | = | My / P = 40.17/586.34 = 69 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(28,21) = 28 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(69,20) = 69 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(28 mm) > 0.05 x 300(15 mm)
and eccenY(69 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00286 | 352.67 | 8.92 | 343.75 | 77.75 | 179 | 13.92 |
| 2 | 226 | 0.00161 | 305.12 | 8.59 | 296.53 | 67.07 | 90 | 6.00 |
| 3 | 226 | 0.00037 | 73.43 | 2.97 | 70.45 | 15.94 | 0 | 0.00 |
| 4 | 226 | -0.00088 | -175.81 | 0.00 | -175.81 | -39.77 | -90 | 3.56 |
| 5 | 226 | -0.00213 | -332.34 | 0.00 | -332.34 | -75.17 | -179 | 13.46 |
| Total | 45.82 | | 36.94 |
| xux | = | 251 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 251/450 = 0.201 |
| Puc | = | 0.201 x 20.00 x 300 x 450 |
| = | 542.95 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 542.95 + (45.82) | = | 588.78 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 251/450 |
| = | 0.232 |
| Muc | = | 542.95 x (0.5 x 450 - 0.232 x 450) = 65.39 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 65.39 + (36.94) |
| = | 102.32 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00258 | 347.28 | 8.92 | 338.36 | 191.34 | 104 | 19.90 |
| 2 | 565 | -0.00156 | -299.82 | 0.00 | -299.82 | -169.55 | -104 | 17.63 |
| Total | 21.79 | | 37.53 |
| xuy | = | 176 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 176/300 = 0.211 |
| Puc | = | 0.211 x 20.00 x 450 x 300 |
| = | 569.53 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 569.53 + (21.79) | = | 591.32 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 176/300 |
| = | 0.244 |
| Muc | = | 569.53 x (0.5 x 300 - 0.244 x 300) = 43.78 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 43.78 + (37.53) |
| = | 81.31 kN-m |
| Pu/Puz | = | 0.377 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.377, | an | = | 1.294 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((16.25/102.32)1.294) + ((40.17/81.31)1.294) |
| = 0.494 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG13 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 877.93 kN |
| MomentX,(Mx) | = | 14.32 kN-m |
| MomentY,(My) | = | 35.65 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C21 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 18.173 kN.m |
| My_MinEccen | = | 17.559 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 18.173 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 35.655 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 14.32/877.93 = 16 mm |
| Actual eccenY | = | My / P = 35.65/877.93 = 41 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(16,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(41,20) = 41 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(41 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00301 | 354.01 | 8.92 | 345.09 | 78.06 | 179 | 13.97 |
| 2 | 226 | 0.00206 | 330.12 | 8.92 | 321.20 | 72.65 | 90 | 6.50 |
| 3 | 226 | 0.00112 | 223.40 | 7.18 | 216.22 | 48.91 | 0 | 0.00 |
| 4 | 226 | 0.00017 | 33.83 | 1.44 | 32.38 | 7.32 | -90 | -0.66 |
| 5 | 226 | -0.00078 | -155.75 | 0.00 | -155.75 | -35.23 | -179 | 6.31 |
| Total | 171.71 | | 26.13 |
| xux | = | 330 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 330/450 = 0.264 |
| Puc | = | 0.264 x 20.00 x 300 x 450 |
| = | 713.81 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 713.81 + (171.71) | = | 885.53 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 330/450 |
| = | 0.306 |
| Muc | = | 713.81 x (0.5 x 450 - 0.306 x 450) = 62.48 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 62.48 + (26.13) |
| = | 88.60 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00279 | 352.08 | 8.92 | 343.16 | 194.05 | 104 | 20.18 |
| 2 | 565 | -0.00041 | -82.08 | 0.00 | -82.08 | -46.41 | -104 | 4.83 |
| Total | 147.64 | | 25.01 |
| xuy | = | 227 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 227/300 = 0.273 |
| Puc | = | 0.273 x 20.00 x 450 x 300 |
| = | 736.59 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 736.59 + (147.64) | = | 884.23 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 227/300 |
| = | 0.315 |
| Muc | = | 736.59 x (0.5 x 300 - 0.315 x 300) = 40.83 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 40.83 + (25.01) |
| = | 65.83 kN-m |
| Pu/Puz | = | 0.564 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.564, | an | = | 1.607 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((18.17/88.60)1.607) + ((35.65/65.83)1.607) |
| = 0.452 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG13 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1150.83 kN |
| MomentX,(Mx) | = | 11.38 kN-m |
| MomentY,(My) | = | 28.95 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C21 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 23.822 kN.m |
| My_MinEccen | = | 23.017 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 23.822 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 28.947 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 11.38/1150.83 = 10 mm |
| Actual eccenY | = | My / P = 28.95/1150.83 = 25 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(10,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(25,20) = 25 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00311 | 354.88 | 8.92 | 345.96 | 78.25 | 179 | 14.01 |
| 2 | 226 | 0.00236 | 340.85 | 8.92 | 331.93 | 75.08 | 90 | 6.72 |
| 3 | 226 | 0.00160 | 304.02 | 8.57 | 295.45 | 66.83 | 0 | 0.00 |
| 4 | 226 | 0.00085 | 169.32 | 5.95 | 163.36 | 36.95 | -90 | -3.31 |
| 5 | 226 | 0.00009 | 18.30 | 0.80 | 17.50 | 3.96 | -179 | -0.71 |
| Total | 261.07 | | 16.71 |
| xux | = | 415 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 415/450 = 0.332 |
| Puc | = | 0.332 x 20.00 x 300 x 450 |
| = | 896.06 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 896.06 + (261.07) | = | 1157.14 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 415/450 |
| = | 0.384 |
| Muc | = | 896.06 x (0.5 x 450 - 0.384 x 450) = 46.98 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 46.98 + (16.71) |
| = | 63.69 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00293 | 353.31 | 8.92 | 344.39 | 194.75 | 104 | 20.25 |
| 2 | 565 | 0.00037 | 73.05 | 2.96 | 70.09 | 39.63 | -104 | -4.12 |
| Total | 234.38 | | 16.13 |
| xuy | = | 284 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 284/300 = 0.340 |
| Puc | = | 0.340 x 20.00 x 450 x 300 |
| = | 918.84 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 918.84 + (234.38) | = | 1153.22 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 284/300 |
| = | 0.393 |
| Muc | = | 918.84 x (0.5 x 300 - 0.393 x 300) = 29.43 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 29.43 + (16.13) |
| = | 45.56 kN-m |
| Pu/Puz | = | 0.739 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.739, | an | = | 1.899 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((23.82/63.69)1.899) + ((28.95/45.56)1.899) |
| = 0.577 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG13 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3375 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1390.20 kN |
| MomentX,(Mx) | = | 5.61 kN-m |
| MomentY,(My) | = | 15.81 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C21 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 30.237 kN.m |
| My_MinEccen | = | 27.804 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 30.237 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 27.804 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 5.61/1390.20 = 4 mm |
| Actual eccenY | = | My / P = 15.81/1390.20 = 11 mm |
| eccenXMin | = | (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(4,22) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(11,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 14 nos. (1583 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00311 | 354.88 | 8.92 | 345.96 | 78.25 | 179 | 14.01 |
| 2 | 226 | 0.00266 | 349.23 | 8.92 | 340.31 | 76.98 | 119 | 9.19 |
| 3 | 226 | 0.00221 | 335.39 | 8.92 | 326.47 | 73.85 | 60 | 4.41 |
| 4 | 226 | 0.00176 | 314.60 | 8.79 | 305.82 | 69.17 | -0 | -0.00 |
| 5 | 226 | 0.00130 | 260.99 | 7.84 | 253.14 | 57.26 | -60 | -3.42 |
| 6 | 226 | 0.00085 | 170.64 | 5.99 | 164.65 | 37.24 | -119 | -4.44 |
| 7 | 226 | 0.00040 | 80.30 | 3.22 | 77.08 | 17.43 | -179 | -3.12 |
| Total | 410.19 | | 16.62 |
| xux | = | 457 mm Puc = C1.fck.B.D |
| ku | = | 1.016 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 1.083 |
| C1 | = | 0.446 x (1 - C3/6) = 0.366 |
| Puc | = | 0.366 x 20.00 x 300 x 450 |
| = | 986.88 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 986.88 + (410.19) | = | 1397.07 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.421 |
| Muc | = | 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 34.93 + (16.62) |
| = | 51.54 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 792 | 0.00287 | 352.75 | 8.92 | 343.83 | 272.20 | 104 | 28.31 |
| 2 | 792 | 0.00068 | 136.19 | 5.04 | 131.15 | 103.83 | -104 | -10.80 |
| Total | 376.03 | | 17.51 |
| xuy | = | 319 mm Puc = C1.fck.B.D |
| ku | = | 1.063 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 0.929 |
| C1 | = | 0.446 x (1 - C3/6) = 0.377 |
| Puc | = | 0.377 x 20.00 x 450 x 300 |
| = | 1017.83 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 1017.83 + (376.03) | = | 1393.86 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.435 |
| Muc | = | 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 19.97 + (17.51) |
| = | 37.48 kN-m |
| Pu/Puz | = | 0.821 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.821, | an | = | 2.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((30.24/51.54)2.000) + ((27.80/37.48)2.000) |
| = 0.894 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG13 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 14 nos. (1583 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG14 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 531.28 kN |
| MomentX,(Mx) | = | 39.57 kN-m |
| MomentY,(My) | = | 13.59 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C22 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 10.998 kN.m |
| My_MinEccen | = | 10.626 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 39.574 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 13.586 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 39.57/531.28 = 74 mm |
| Actual eccenY | = | My / P = 13.59/531.28 = 26 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(74,21) = 74 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(26,20) = 26 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(74 mm) > 0.05 x 300(15 mm)
and eccenY(26 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00282 | 352.34 | 8.92 | 343.42 | 77.68 | 179 | 13.90 |
| 2 | 226 | 0.00150 | 294.53 | 8.37 | 286.16 | 64.73 | 90 | 5.79 |
| 3 | 226 | 0.00018 | 36.30 | 1.55 | 34.75 | 7.86 | 0 | 0.00 |
| 4 | 226 | -0.00114 | -227.71 | 0.00 | -227.71 | -51.51 | -90 | 4.61 |
| 5 | 226 | -0.00246 | -344.05 | 0.00 | -344.05 | -77.82 | -179 | 13.93 |
| Total | 20.94 | | 38.24 |
| xux | = | 237 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 237/450 = 0.190 |
| Puc | = | 0.190 x 20.00 x 300 x 450 |
| = | 512.58 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 512.58 + (20.94) | = | 533.52 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 237/450 |
| = | 0.219 |
| Muc | = | 512.58 x (0.5 x 450 - 0.219 x 450) = 64.73 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 64.73 + (38.24) |
| = | 102.97 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00251 | 345.41 | 8.92 | 336.49 | 190.28 | 104 | 19.79 |
| 2 | 565 | -0.00196 | -326.18 | 0.00 | -326.18 | -184.45 | -104 | 19.18 |
| Total | 5.83 | | 38.97 |
| xuy | = | 163 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 163/300 = 0.195 |
| Puc | = | 0.195 x 20.00 x 450 x 300 |
| = | 527.77 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 527.77 + (5.83) | = | 533.60 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 163/300 |
| = | 0.226 |
| Muc | = | 527.77 x (0.5 x 300 - 0.226 x 300) = 43.40 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 43.40 + (38.97) |
| = | 82.37 kN-m |
| Pu/Puz | = | 0.341 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.341, | an | = | 1.235 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((39.57/102.97)1.235) + ((13.59/82.37)1.235) |
| = 0.415 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG14 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1049.43 kN |
| MomentX,(Mx) | = | 31.95 kN-m |
| MomentY,(My) | = | 12.77 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C22 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 21.723 kN.m |
| My_MinEccen | = | 20.989 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 31.951 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 20.989 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 31.95/1049.43 = 30 mm |
| Actual eccenY | = | My / P = 12.77/1049.43 = 12 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(30,21) = 30 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(12,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(30 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00308 | 354.56 | 8.92 | 345.64 | 78.18 | 179 | 13.99 |
| 2 | 226 | 0.00225 | 336.96 | 8.92 | 328.04 | 74.20 | 90 | 6.64 |
| 3 | 226 | 0.00143 | 285.19 | 8.19 | 277.00 | 62.66 | 0 | 0.00 |
| 4 | 226 | 0.00060 | 120.18 | 4.55 | 115.63 | 26.15 | -90 | -2.34 |
| 5 | 226 | -0.00022 | -44.82 | 0.00 | -44.82 | -10.14 | -179 | 1.81 |
| Total | 231.05 | | 20.11 |
| xux | = | 380 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 380/450 = 0.304 |
| Puc | = | 0.304 x 20.00 x 300 x 450 |
| = | 820.13 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 820.13 + (231.05) | = | 1051.18 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 380/450 |
| = | 0.351 |
| Muc | = | 820.13 x (0.5 x 450 - 0.351 x 450) = 54.99 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 54.99 + (20.11) |
| = | 75.10 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00289 | 352.91 | 8.92 | 343.99 | 194.52 | 104 | 20.23 |
| 2 | 565 | 0.00011 | 22.67 | 0.98 | 21.68 | 12.26 | -104 | -1.28 |
| Total | 206.78 | | 18.95 |
| xuy | = | 263 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 263/300 = 0.315 |
| Puc | = | 0.315 x 20.00 x 450 x 300 |
| = | 850.50 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 850.50 + (206.78) | = | 1057.28 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 263/300 |
| = | 0.364 |
| Muc | = | 850.50 x (0.5 x 300 - 0.364 x 300) = 34.70 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 34.70 + (18.95) |
| = | 53.66 kN-m |
| Pu/Puz | = | 0.674 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.674, | an | = | 1.790 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((31.95/75.10)1.790) + ((20.99/53.66)1.790) |
| = 0.403 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG14 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1584.54 kN |
| MomentX,(Mx) | = | 34.96 kN-m |
| MomentY,(My) | = | 10.12 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C22 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 32.800 kN.m |
| My_MinEccen | = | 31.691 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 34.963 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 31.691 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 34.96/1584.54 = 22 mm |
| Actual eccenY | = | My / P = 10.12/1584.54 = 6 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(22,21) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(6,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 20 nos. (2262 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00308 | 354.63 | 8.92 | 345.71 | 78.20 | 179 | 14.00 |
| 2 | 226 | 0.00279 | 352.06 | 8.92 | 343.14 | 77.62 | 139 | 10.81 |
| 3 | 226 | 0.00250 | 345.02 | 8.92 | 336.10 | 76.02 | 99 | 7.56 |
| 4 | 226 | 0.00220 | 335.19 | 8.92 | 326.27 | 73.80 | 60 | 4.40 |
| 5 | 226 | 0.00191 | 324.15 | 8.90 | 315.25 | 71.31 | 20 | 1.42 |
| 6 | 226 | 0.00162 | 305.40 | 8.59 | 296.81 | 67.14 | -20 | -1.34 |
| 7 | 226 | 0.00132 | 264.59 | 7.90 | 256.69 | 58.06 | -60 | -3.46 |
| 8 | 226 | 0.00103 | 205.92 | 6.82 | 199.10 | 45.04 | -99 | -4.48 |
| 9 | 226 | 0.00074 | 147.25 | 5.36 | 141.90 | 32.10 | -139 | -4.47 |
| 10 | 226 | 0.00044 | 88.59 | 3.51 | 85.07 | 19.24 | -179 | -3.44 |
| Total | 598.52 | | 20.99 |
| xux | = | 464 mm Puc = C1.fck.B.D |
| ku | = | 1.031 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 1.027 |
| C1 | = | 0.446 x (1 - C3/6) = 0.370 |
| Puc | = | 0.370 x 20.00 x 300 x 450 |
| = | 998.00 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 998.00 + (598.52) | = | 1596.52 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.426 |
| Muc | = | 998.00 x (0.5 x 450 - 0.426 x 450) = 33.14 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 33.14 + (20.99) |
| = | 54.13 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1131 | 0.00283 | 352.39 | 8.92 | 343.47 | 388.46 | 104 | 40.40 |
| 2 | 1131 | 0.00074 | 148.58 | 5.40 | 143.19 | 161.94 | -104 | -16.84 |
| Total | 550.40 | | 23.56 |
| xuy | = | 328 mm Puc = C1.fck.B.D |
| ku | = | 1.094 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 0.843 |
| C1 | = | 0.446 x (1 - C3/6) = 0.383 |
| Puc | = | 0.383 x 20.00 x 450 x 300 |
| = | 1034.93 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 1034.93 + (550.40) | = | 1585.32 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.442 |
| Muc | = | 1034.93 x (0.5 x 300 - 0.442 x 300) = 18.14 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 18.14 + (23.56) |
| = | 41.69 kN-m |
| Pu/Puz | = | 0.835 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.835, | an | = | 2.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((34.96/54.13)2.000) + ((31.69/41.69)2.000) |
| = 0.995 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG14 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 20 nos. (2262 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 2147.07 kN |
| MomentX,(Mx) | = | 42.32 kN-m |
| MomentY,(My) | = | 6.52 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C22 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 44.444 kN.m |
| My_MinEccen | = | 42.941 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 44.444 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 42.941 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 42.32/2147.07 = 20 mm |
| Actual eccenY | = | My / P = 6.52/2147.07 = 3 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(20,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 628 | 0.00305 | 354.37 | 8.92 | 345.45 | 217.05 | 175 | 37.98 |
| 2 | 628 | 0.00262 | 348.29 | 8.92 | 339.37 | 213.23 | 117 | 24.88 |
| 3 | 628 | 0.00219 | 334.83 | 8.92 | 325.91 | 204.78 | 58 | 11.95 |
| 4 | 628 | 0.00176 | 315.00 | 8.79 | 306.20 | 192.39 | 0 | 0.00 |
| 5 | 628 | 0.00133 | 266.56 | 7.93 | 258.63 | 162.50 | -58 | -9.48 |
| 6 | 628 | 0.00090 | 180.52 | 6.23 | 174.29 | 109.51 | -117 | -12.78 |
| 7 | 628 | 0.00047 | 94.49 | 3.72 | 90.77 | 57.03 | -175 | -9.98 |
| Total | 1156.49 | | 42.57 |
| xux | = | 464 mm Puc = C1.fck.B.D |
| ku | = | 1.031 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 1.027 |
| C1 | = | 0.446 x (1 - C3/6) = 0.370 |
| Puc | = | 0.370 x 20.00 x 300 x 450 |
| = | 998.00 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 998.00 + (1156.49) | = | 2154.49 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.426 |
| Muc | = | 998.00 x (0.5 x 450 - 0.426 x 450) = 33.14 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 33.14 + (42.57) |
| = | 75.71 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 2199 | 0.00275 | 351.60 | 8.92 | 342.68 | 753.59 | 100 | 75.36 |
| 2 | 2199 | 0.00084 | 167.52 | 5.91 | 161.61 | 355.41 | -100 | -35.54 |
| Total | 1109.00 | | 39.82 |
| xuy | = | 338 mm Puc = C1.fck.B.D |
| ku | = | 1.125 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 0.769 |
| C1 | = | 0.446 x (1 - C3/6) = 0.389 |
| Puc | = | 0.389 x 20.00 x 450 x 300 |
| = | 1049.78 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 1049.78 + (1109.00) | = | 2158.77 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.447 |
| Muc | = | 1049.78 x (0.5 x 300 - 0.447 x 300) = 16.55 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 16.55 + (39.82) |
| = | 56.36 kN-m |
| Pu/Puz | = | 0.844 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.844, | an | = | 2.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((44.44/75.71)2.000) + ((42.94/56.36)2.000) |
| = 0.925 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 20 = 320 mm
iii) 300 mm
Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties 8 @ 300 mm c/c
SUMMARY :
CG14 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3375 mm |
From analysis results,loads on column
| Axial load,(P) | = | 2759.80 kN |
| MomentX,(Mx) | = | 26.29 kN-m |
| MomentY,(My) | = | 2.48 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C22 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 60.026 kN.m |
| My_MinEccen | = | 55.196 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 60.026 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 55.196 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 26.29/2759.80 = 10 mm |
| Actual eccenY | = | My / P = 2.48/2759.80 = 1 mm |
| eccenXMin | = | (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(10,22) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 14 nos. (6872 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 982 | 0.00306 | 354.45 | 8.92 | 345.53 | 339.22 | 173 | 58.52 |
| 2 | 982 | 0.00263 | 348.39 | 8.92 | 339.47 | 333.27 | 115 | 38.33 |
| 3 | 982 | 0.00219 | 334.79 | 8.92 | 325.87 | 319.92 | 58 | 18.40 |
| 4 | 982 | 0.00176 | 314.60 | 8.79 | 305.82 | 300.24 | 0 | 0.00 |
| 5 | 982 | 0.00132 | 264.27 | 7.89 | 256.37 | 251.69 | -58 | -14.47 |
| 6 | 982 | 0.00089 | 177.20 | 6.15 | 171.05 | 167.93 | -115 | -19.31 |
| 7 | 982 | 0.00045 | 90.14 | 3.57 | 86.57 | 84.99 | -173 | -14.66 |
| Total | 1797.27 | | 66.79 |
| xux | = | 457 mm Puc = C1.fck.B.D |
| ku | = | 1.016 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 1.083 |
| C1 | = | 0.446 x (1 - C3/6) = 0.366 |
| Puc | = | 0.366 x 20.00 x 300 x 450 |
| = | 986.88 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 986.88 + (1797.27) | = | 2784.14 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.421 |
| Muc | = | 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 34.93 + (66.79) |
| = | 101.72 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 3436 | 0.00274 | 351.41 | 8.92 | 342.49 | 1176.84 | 98 | 114.74 |
| 2 | 3436 | 0.00084 | 167.08 | 5.90 | 161.19 | 553.86 | -98 | -54.00 |
| Total | 1730.70 | | 60.74 |
| xuy | = | 333 mm Puc = C1.fck.B.D |
| ku | = | 1.109 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 0.805 |
| C1 | = | 0.446 x (1 - C3/6) = 0.386 |
| Puc | = | 0.386 x 20.00 x 450 x 300 |
| = | 1042.61 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 1042.61 + (1730.70) | = | 2773.31 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.445 |
| Muc | = | 1042.61 x (0.5 x 300 - 0.445 x 300) = 17.31 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 17.31 + (60.74) |
| = | 78.05 kN-m |
| Pu/Puz | = | 0.838 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.838, | an | = | 2.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((60.03/101.72)2.000) + ((55.20/78.05)2.000) |
| = 0.848 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 25 = 400 mm
iii) 300 mm
Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties 8 @ 300 mm c/c
SUMMARY :
CG14 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 14 nos. (6872 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG15 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 604.88 kN |
| MomentX,(Mx) | = | 13.89 kN-m |
| MomentY,(My) | = | 9.20 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C23 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 12.521 kN.m |
| My_MinEccen | = | 12.098 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 13.894 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 12.098 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 13.89/604.88 = 23 mm |
| Actual eccenY | = | My / P = 9.20/604.88 = 15 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(23,21) = 23 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(15,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(23 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00287 | 352.79 | 8.92 | 343.87 | 77.78 | 179 | 13.92 |
| 2 | 226 | 0.00165 | 308.08 | 8.65 | 299.43 | 67.73 | 90 | 6.06 |
| 3 | 226 | 0.00043 | 86.30 | 3.43 | 82.87 | 18.74 | 0 | 0.00 |
| 4 | 226 | -0.00079 | -157.81 | 0.00 | -157.81 | -35.70 | -90 | 3.19 |
| 5 | 226 | -0.00201 | -328.09 | 0.00 | -328.09 | -74.21 | -179 | 13.28 |
| Total | 54.34 | | 36.46 |
| xux | = | 257 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 257/450 = 0.205 |
| Puc | = | 0.205 x 20.00 x 300 x 450 |
| = | 554.34 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 554.34 + (54.34) | = | 608.69 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 257/450 |
| = | 0.237 |
| Muc | = | 554.34 x (0.5 x 450 - 0.237 x 450) = 65.54 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 65.54 + (36.46) |
| = | 102.01 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00260 | 347.74 | 8.92 | 338.82 | 191.60 | 104 | 19.93 |
| 2 | 565 | -0.00146 | -290.43 | 0.00 | -290.43 | -164.23 | -104 | 17.08 |
| Total | 27.36 | | 37.01 |
| xuy | = | 179 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 179/300 = 0.215 |
| Puc | = | 0.215 x 20.00 x 450 x 300 |
| = | 580.92 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 580.92 + (27.36) | = | 608.29 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 179/300 |
| = | 0.249 |
| Muc | = | 580.92 x (0.5 x 300 - 0.249 x 300) = 43.81 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 43.81 + (37.01) |
| = | 80.82 kN-m |
| Pu/Puz | = | 0.389 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.389, | an | = | 1.314 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((13.89/102.01)1.314) + ((12.10/80.82)1.314) |
| = 0.155 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG15 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1214.02 kN |
| MomentX,(Mx) | = | 14.42 kN-m |
| MomentY,(My) | = | 8.59 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C23 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 25.130 kN.m |
| My_MinEccen | = | 24.280 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 25.130 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 24.280 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 14.42/1214.02 = 12 mm |
| Actual eccenY | = | My / P = 8.59/1214.02 = 7 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(12,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(7,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00313 | 355.04 | 8.92 | 346.12 | 78.29 | 179 | 14.01 |
| 2 | 226 | 0.00241 | 342.85 | 8.92 | 333.93 | 75.53 | 90 | 6.76 |
| 3 | 226 | 0.00169 | 310.67 | 8.71 | 301.96 | 68.30 | 0 | 0.00 |
| 4 | 226 | 0.00097 | 195.00 | 6.58 | 188.42 | 42.62 | -90 | -3.81 |
| 5 | 226 | 0.00026 | 51.28 | 2.14 | 49.14 | 11.12 | -179 | -1.99 |
| Total | 275.86 | | 14.97 |
| xux | = | 436 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 436/450 = 0.349 |
| Puc | = | 0.349 x 20.00 x 300 x 450 |
| = | 941.63 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 941.63 + (275.86) | = | 1217.49 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 436/450 |
| = | 0.403 |
| Muc | = | 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 41.10 + (14.97) |
| = | 56.07 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00296 | 353.58 | 8.92 | 344.66 | 194.90 | 104 | 20.27 |
| 2 | 565 | 0.00054 | 107.33 | 4.14 | 103.19 | 58.35 | -104 | -6.07 |
| Total | 253.25 | | 14.20 |
| xuy | = | 300 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 300/300 = 0.360 |
| Puc | = | 0.360 x 20.00 x 450 x 300 |
| = | 972.00 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 972.00 + (253.25) | = | 1225.25 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 300/300 |
| = | 0.416 |
| Muc | = | 972.00 x (0.5 x 300 - 0.416 x 300) = 24.49 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 24.49 + (14.20) |
| = | 38.70 kN-m |
| Pu/Puz | = | 0.780 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.780, | an | = | 1.966 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((25.13/56.07)1.966) + ((24.28/38.70)1.966) |
| = 0.606 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG15 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1803.60 kN |
| MomentX,(Mx) | = | 10.42 kN-m |
| MomentY,(My) | = | 7.70 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C23 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 37.334 kN.m |
| My_MinEccen | = | 36.072 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 37.334 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 36.072 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 10.42/1803.60 = 6 mm |
| Actual eccenY | = | My / P = 7.70/1803.60 = 4 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(6,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(4,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00313 | 355.01 | 8.92 | 346.09 | 139.17 | 177 | 24.63 |
| 2 | 402 | 0.00273 | 351.11 | 8.92 | 342.19 | 137.60 | 126 | 17.40 |
| 3 | 402 | 0.00234 | 340.23 | 8.92 | 331.31 | 133.23 | 76 | 10.11 |
| 4 | 402 | 0.00195 | 325.78 | 8.91 | 316.87 | 127.42 | 25 | 3.22 |
| 5 | 402 | 0.00155 | 299.44 | 8.48 | 290.96 | 117.00 | -25 | -2.96 |
| 6 | 402 | 0.00116 | 232.00 | 7.35 | 224.65 | 90.34 | -76 | -6.85 |
| 7 | 402 | 0.00077 | 153.33 | 5.53 | 147.81 | 59.44 | -126 | -7.51 |
| 8 | 402 | 0.00037 | 74.67 | 3.02 | 71.65 | 28.81 | -177 | -5.10 |
| Total | 833.01 | | 32.93 |
| xux | = | 450 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 450/450 = 0.360 |
| Puc | = | 0.360 x 20.00 x 300 x 450 |
| = | 972.00 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 972.00 + (833.01) | = | 1805.01 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 450/450 |
| = | 0.416 |
| Muc | = | 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 36.74 + (32.93) |
| = | 69.67 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1608 | 0.00283 | 352.39 | 8.92 | 343.47 | 552.46 | 102 | 56.35 |
| 2 | 1608 | 0.00073 | 146.64 | 5.34 | 141.30 | 227.28 | -102 | -23.18 |
| Total | 779.74 | | 33.17 |
| xuy | = | 323 mm Puc = C1.fck.B.D |
| ku | = | 1.078 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 0.884 |
| C1 | = | 0.446 x (1 - C3/6) = 0.380 |
| Puc | = | 0.380 x 20.00 x 450 x 300 |
| = | 1026.69 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 1026.69 + (779.74) | = | 1806.43 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.438 |
| Muc | = | 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 19.02 + (33.17) |
| = | 52.19 kN-m |
| Pu/Puz | = | 0.825 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.825, | an | = | 2.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((37.33/69.67)2.000) + ((36.07/52.19)2.000) |
| = 0.765 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG15 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 2364.75 kN |
| MomentX,(Mx) | = | 3.76 kN-m |
| MomentY,(My) | = | 6.67 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C23 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 48.950 kN.m |
| My_MinEccen | = | 47.295 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 48.950 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 47.295 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 3.76/2364.75 = 2 mm |
| Actual eccenY | = | My / P = 6.67/2364.75 = 3 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 12 nos. (5890 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 982 | 0.00307 | 354.53 | 8.92 | 345.61 | 339.30 | 173 | 58.53 |
| 2 | 982 | 0.00251 | 345.33 | 8.92 | 336.41 | 330.27 | 104 | 34.18 |
| 3 | 982 | 0.00195 | 325.74 | 8.91 | 316.82 | 311.04 | 35 | 10.73 |
| 4 | 982 | 0.00138 | 276.48 | 8.07 | 268.41 | 263.51 | -35 | -9.09 |
| 5 | 982 | 0.00082 | 163.87 | 5.81 | 158.06 | 155.17 | -104 | -16.06 |
| 6 | 982 | 0.00026 | 51.26 | 2.14 | 49.12 | 48.22 | -173 | -8.32 |
| Total | 1447.52 | | 69.97 |
| xux | = | 429 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 429/450 = 0.343 |
| Puc | = | 0.343 x 20.00 x 300 x 450 |
| = | 926.44 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 926.44 + (1447.52) | = | 2373.95 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 429/450 |
| = | 0.397 |
| Muc | = | 926.44 x (0.5 x 450 - 0.397 x 450) = 43.15 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 43.15 + (69.97) |
| = | 113.12 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 2945 | 0.00286 | 352.71 | 8.92 | 343.79 | 1012.54 | 98 | 98.72 |
| 2 | 2945 | 0.00065 | 129.89 | 4.85 | 125.03 | 368.25 | -98 | -35.90 |
| Total | 1380.80 | | 62.82 |
| xuy | = | 305 mm Puc = C1.fck.B.D |
| ku | = | 1.016 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 1.083 |
| C1 | = | 0.446 x (1 - C3/6) = 0.366 |
| Puc | = | 0.366 x 20.00 x 450 x 300 |
| = | 986.88 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 986.88 + (1380.80) | = | 2367.67 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.421 |
| Muc | = | 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 23.28 + (62.82) |
| = | 86.10 kN-m |
| Pu/Puz | = | 0.789 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.789, | an | = | 1.982 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((48.95/113.12)1.982) + ((47.30/86.10)1.982) |
| = 0.495 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 25 = 400 mm
iii) 300 mm
Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties 8 @ 300 mm c/c
SUMMARY :
CG15 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 12 nos. (5890 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3375 mm |
From analysis results,loads on column
| Axial load,(P) | = | 2859.01 kN |
| MomentX,(Mx) | = | 0.84 kN-m |
| MomentY,(My) | = | 4.04 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C23 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 62.184 kN.m |
| My_MinEccen | = | 57.180 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 62.184 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 57.180 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 0.84/2859.01 = 0 mm |
| Actual eccenY | = | My / P = 4.04/2859.01 = 1 mm |
| eccenXMin | = | (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(0,22) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 10 nos. (8042 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1608 | 0.00304 | 354.24 | 8.92 | 345.32 | 555.45 | 169 | 93.87 |
| 2 | 1608 | 0.00234 | 340.38 | 8.92 | 331.46 | 533.14 | 85 | 45.05 |
| 3 | 1608 | 0.00165 | 307.87 | 8.64 | 299.23 | 481.30 | 0 | 0.00 |
| 4 | 1608 | 0.00095 | 190.70 | 6.48 | 184.23 | 296.33 | -85 | -25.04 |
| 5 | 1608 | 0.00026 | 51.65 | 2.16 | 49.50 | 79.62 | -169 | -13.46 |
| Total | 1945.85 | | 100.43 |
| xux | = | 425 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 425/450 = 0.340 |
| Puc | = | 0.340 x 20.00 x 300 x 450 |
| = | 918.84 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 918.84 + (1945.85) | = | 2864.69 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 425/450 |
| = | 0.393 |
| Muc | = | 918.84 x (0.5 x 450 - 0.393 x 450) = 44.14 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 44.14 + (100.43) |
| = | 144.57 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 4021 | 0.00285 | 352.56 | 8.92 | 343.64 | 1381.85 | 94 | 129.89 |
| 2 | 4021 | 0.00065 | 130.67 | 4.88 | 125.79 | 505.83 | -94 | -47.55 |
| Total | 1887.69 | | 82.35 |
| xuy | = | 300 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 300/300 = 0.360 |
| Puc | = | 0.360 x 20.00 x 450 x 300 |
| = | 972.00 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 972.00 + (1887.69) | = | 2859.69 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 300/300 |
| = | 0.416 |
| Muc | = | 972.00 x (0.5 x 300 - 0.416 x 300) = 24.49 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 24.49 + (82.35) |
| = | 106.84 kN-m |
| Pu/Puz | = | 0.784 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.784, | an | = | 1.974 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((62.18/144.57)1.974) + ((57.18/106.84)1.974) |
| = 0.480 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 32 = 512 mm
iii) 300 mm
Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties 8 @ 300 mm c/c
SUMMARY :
CG15 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 10 nos. (8042 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG16 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 650.80 kN |
| MomentX,(Mx) | = | 62.68 kN-m |
| MomentY,(My) | = | 2.05 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C24 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 13.472 kN.m |
| My_MinEccen | = | 13.016 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 62.678 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 13.016 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 62.68/650.80 = 96 mm |
| Actual eccenY | = | My / P = 2.05/650.80 = 3 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(96,21) = 96 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(96 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00290 | 353.04 | 8.92 | 344.12 | 77.84 | 179 | 13.93 |
| 2 | 226 | 0.00174 | 313.36 | 8.77 | 304.59 | 68.90 | 90 | 6.17 |
| 3 | 226 | 0.00057 | 114.38 | 4.37 | 110.01 | 24.88 | 0 | 0.00 |
| 4 | 226 | -0.00059 | -118.57 | 0.00 | -118.57 | -26.82 | -90 | 2.40 |
| 5 | 226 | -0.00176 | -314.66 | 0.00 | -314.66 | -71.17 | -179 | 12.74 |
| Total | 73.62 | | 35.24 |
| xux | = | 269 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 269/450 = 0.215 |
| Puc | = | 0.215 x 20.00 x 300 x 450 |
| = | 580.92 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 580.92 + (73.62) | = | 654.54 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 269/450 |
| = | 0.249 |
| Muc | = | 580.92 x (0.5 x 450 - 0.249 x 450) = 65.71 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 65.71 + (35.24) |
| = | 100.95 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00264 | 348.61 | 8.92 | 339.69 | 192.09 | 104 | 19.98 |
| 2 | 565 | -0.00127 | -254.23 | 0.00 | -254.23 | -143.76 | -104 | 14.95 |
| Total | 48.33 | | 34.93 |
| xuy | = | 186 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 186/300 = 0.224 |
| Puc | = | 0.224 x 20.00 x 450 x 300 |
| = | 603.70 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 603.70 + (48.33) | = | 652.03 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 186/300 |
| = | 0.258 |
| Muc | = | 603.70 x (0.5 x 300 - 0.258 x 300) = 43.76 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 43.76 + (34.93) |
| = | 78.69 kN-m |
| Pu/Puz | = | 0.418 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.418, | an | = | 1.363 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((62.68/100.95)1.363) + ((13.02/78.69)1.363) |
| = 0.608 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG16 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1301.16 kN |
| MomentX,(Mx) | = | 54.20 kN-m |
| MomentY,(My) | = | 1.06 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C24 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 26.934 kN.m |
| My_MinEccen | = | 26.023 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 54.200 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 26.023 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 54.20/1301.16 = 42 mm |
| Actual eccenY | = | My / P = 1.06/1301.16 = 1 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(42,21) = 42 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(42 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 8 nos. (1608 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 402 | 0.00311 | 354.85 | 8.92 | 345.93 | 139.11 | 177 | 24.62 |
| 2 | 402 | 0.00215 | 333.08 | 8.92 | 324.16 | 130.35 | 59 | 7.69 |
| 3 | 402 | 0.00118 | 236.50 | 7.43 | 229.07 | 92.11 | -59 | -5.43 |
| 4 | 402 | 0.00022 | 43.91 | 1.85 | 42.06 | 16.91 | -177 | -2.99 |
| Total | 378.48 | | 23.88 |
| xux | = | 429 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 429/450 = 0.343 |
| Puc | = | 0.343 x 20.00 x 300 x 450 |
| = | 926.44 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 926.44 + (378.48) | = | 1304.92 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 429/450 |
| = | 0.397 |
| Muc | = | 926.44 x (0.5 x 450 - 0.397 x 450) = 43.15 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 43.15 + (23.88) |
| = | 67.03 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 804 | 0.00293 | 353.30 | 8.92 | 344.38 | 276.96 | 102 | 28.25 |
| 2 | 804 | 0.00051 | 102.67 | 3.99 | 98.68 | 79.36 | -102 | -8.09 |
| Total | 356.32 | | 20.16 |
| xuy | = | 295 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 295/300 = 0.354 |
| Puc | = | 0.354 x 20.00 x 450 x 300 |
| = | 956.81 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 956.81 + (356.32) | = | 1313.14 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 295/300 |
| = | 0.410 |
| Muc | = | 956.81 x (0.5 x 300 - 0.410 x 300) = 25.98 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 25.98 + (20.16) |
| = | 46.13 kN-m |
| Pu/Puz | = | 0.765 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.765, | an | = | 1.941 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((54.20/67.03)1.941) + ((26.02/46.13)1.941) |
| = 0.991 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 16 = 256 mm
iii) 300 mm
Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties 8 @ 250 mm c/c
SUMMARY :
CG16 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 8 nos. (1608 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor3 - Default Level at 11.700 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 1973.48 kN |
| MomentX,(Mx) | = | 42.96 kN-m |
| MomentY,(My) | = | 1.21 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C24 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 40.851 kN.m |
| My_MinEccen | = | 39.470 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 42.961 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 39.470 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 42.96/1973.48 = 22 mm |
| Actual eccenY | = | My / P = 1.21/1973.48 = 1 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(22,21) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 12 nos. (3770 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 628 | 0.00305 | 354.37 | 8.92 | 345.45 | 217.05 | 175 | 37.98 |
| 2 | 628 | 0.00254 | 346.07 | 8.92 | 337.15 | 211.84 | 105 | 22.24 |
| 3 | 628 | 0.00202 | 328.51 | 8.92 | 319.59 | 200.81 | 35 | 7.03 |
| 4 | 628 | 0.00150 | 294.84 | 8.37 | 286.47 | 180.00 | -35 | -6.30 |
| 5 | 628 | 0.00099 | 197.73 | 6.64 | 191.09 | 120.06 | -105 | -12.61 |
| 6 | 628 | 0.00047 | 94.49 | 3.72 | 90.77 | 57.03 | -175 | -9.98 |
| Total | 986.79 | | 38.37 |
| xux | = | 464 mm Puc = C1.fck.B.D |
| ku | = | 1.031 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 1.027 |
| C1 | = | 0.446 x (1 - C3/6) = 0.370 |
| Puc | = | 0.370 x 20.00 x 300 x 450 |
| = | 998.00 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 998.00 + (986.79) | = | 1984.79 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.426 |
| Muc | = | 998.00 x (0.5 x 450 - 0.426 x 450) = 33.14 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 33.14 + (38.37) |
| = | 71.51 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1885 | 0.00277 | 351.88 | 8.92 | 342.96 | 646.47 | 100 | 64.65 |
| 2 | 1885 | 0.00081 | 162.19 | 5.77 | 156.42 | 294.84 | -100 | -29.48 |
| Total | 941.31 | | 35.16 |
| xuy | = | 333 mm Puc = C1.fck.B.D |
| ku | = | 1.109 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 0.805 |
| C1 | = | 0.446 x (1 - C3/6) = 0.386 |
| Puc | = | 0.386 x 20.00 x 450 x 300 |
| = | 1042.61 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 1042.61 + (941.31) | = | 1983.92 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.445 |
| Muc | = | 1042.61 x (0.5 x 300 - 0.445 x 300) = 17.31 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 17.31 + (35.16) |
| = | 52.48 kN-m |
| Pu/Puz | = | 0.838 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.838, | an | = | 2.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((42.96/71.51)2.000) + ((39.47/52.48)2.000) |
| = 0.927 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 20 = 320 mm
iii) 300 mm
Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties 8 @ 300 mm c/c
SUMMARY :
CG16 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 12 nos. (3770 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor2 - Default Level at 7.800 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 2850 mm |
From analysis results,loads on column
| Axial load,(P) | = | 2598.35 kN |
| MomentX,(Mx) | = | 59.23 kN-m |
| MomentY,(My) | = | 4.06 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C24 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 53.786 kN.m |
| My_MinEccen | = | 51.967 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 59.228 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 51.967 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 59.23/2598.35 = 23 mm |
| Actual eccenY | = | My / P = 4.06/2598.35 = 2 mm |
| eccenXMin | = | (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm |
| eccenXMin | = | max(21,20)= 21 mm |
| eccenYMin | = | (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm |
| eccenYMin | = | max(16,20) = 21 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(23,21) = 23 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(2,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(23 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 14 nos. (6872 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 982 | 0.00307 | 354.53 | 8.92 | 345.61 | 339.30 | 173 | 58.53 |
| 2 | 982 | 0.00260 | 347.75 | 8.92 | 338.83 | 332.64 | 115 | 38.25 |
| 3 | 982 | 0.00213 | 332.63 | 8.92 | 323.71 | 317.80 | 58 | 18.27 |
| 4 | 982 | 0.00166 | 308.82 | 8.67 | 300.15 | 294.67 | 0 | 0.00 |
| 5 | 982 | 0.00119 | 238.94 | 7.47 | 231.47 | 227.24 | -58 | -13.07 |
| 6 | 982 | 0.00073 | 145.10 | 5.30 | 139.80 | 137.25 | -115 | -15.78 |
| 7 | 982 | 0.00026 | 51.26 | 2.14 | 49.12 | 48.22 | -173 | -8.32 |
| Total | 1697.13 | | 77.89 |
| xux | = | 429 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 429/450 = 0.343 |
| Puc | = | 0.343 x 20.00 x 300 x 450 |
| = | 926.44 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 926.44 + (1697.13) | = | 2623.57 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 429/450 |
| = | 0.397 |
| Muc | = | 926.44 x (0.5 x 450 - 0.397 x 450) = 43.15 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 43.15 + (77.89) |
| = | 121.04 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 3436 | 0.00285 | 352.61 | 8.92 | 343.69 | 1180.96 | 98 | 115.14 |
| 2 | 3436 | 0.00067 | 133.43 | 4.96 | 128.47 | 441.45 | -98 | -43.04 |
| Total | 1622.41 | | 72.10 |
| xuy | = | 307 mm Puc = C1.fck.B.D |
| ku | = | 1.023 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 1.055 |
| C1 | = | 0.446 x (1 - C3/6) = 0.368 |
| Puc | = | 0.368 x 20.00 x 450 x 300 |
| = | 992.55 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 992.55 + (1622.41) | = | 2614.96 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.424 |
| Muc | = | 992.55 x (0.5 x 300 - 0.424 x 300) = 22.68 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 22.68 + (72.10) |
| = | 94.78 kN-m |
| Pu/Puz | = | 0.789 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.789, | an | = | 1.982 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((59.23/121.04)1.982) + ((51.97/94.78)1.982) |
| = 0.546 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 25 = 400 mm
iii) 300 mm
Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties 8 @ 300 mm c/c
SUMMARY :
CG16 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 14 nos. (6872 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor1 - Default Level at 3.900 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3375 mm |
From analysis results,loads on column
| Axial load,(P) | = | 3348.72 kN |
| MomentX,(Mx) | = | 50.85 kN-m |
| MomentY,(My) | = | 3.56 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C24 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 72.835 kN.m |
| My_MinEccen | = | 66.974 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 72.835 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 66.974 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 50.85/3348.72 = 15 mm |
| Actual eccenY | = | My / P = 3.56/3348.72 = 1 mm |
| eccenXMin | = | (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(15,22) = 22 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 12 nos. (9651 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 1608 | 0.00305 | 354.34 | 8.92 | 345.42 | 555.61 | 169 | 93.90 |
| 2 | 1608 | 0.00251 | 345.31 | 8.92 | 336.39 | 541.08 | 101 | 54.87 |
| 3 | 1608 | 0.00196 | 326.45 | 8.92 | 317.53 | 510.75 | 34 | 17.26 |
| 4 | 1608 | 0.00142 | 284.44 | 8.18 | 276.26 | 444.36 | -34 | -15.02 |
| 5 | 1608 | 0.00088 | 175.89 | 6.12 | 169.77 | 273.07 | -101 | -27.69 |
| 6 | 1608 | 0.00034 | 67.34 | 2.75 | 64.59 | 103.89 | -169 | -17.56 |
| Total | 2428.77 | | 105.76 |
| xux | = | 436 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 436/450 = 0.349 |
| Puc | = | 0.349 x 20.00 x 300 x 450 |
| = | 941.63 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 941.63 + (2428.77) | = | 3370.39 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 436/450 |
| = | 0.403 |
| Muc | = | 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 41.10 + (105.76) |
| = | 146.86 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 4825 | 0.00278 | 352.00 | 8.92 | 343.08 | 1655.51 | 94 | 155.62 |
| 2 | 4825 | 0.00076 | 151.09 | 5.47 | 145.62 | 702.69 | -94 | -66.05 |
| Total | 2358.20 | | 89.57 |
| xuy | = | 314 mm Puc = C1.fck.B.D |
| ku | = | 1.047 |
| C3 | = | (8/7) x [4/(7.ku - 3)]2 = 0.976 |
| C1 | = | 0.446 x (1 - C3/6) = 0.373 |
| Puc | = | 0.373 x 20.00 x 450 x 300 |
| = | 1008.29 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 1008.29 + (2358.20) | = | 3366.49 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | (0.5f - C3/7)/(1 - C3/6) = 0.431 |
| Muc | = | 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 20.99 + (89.57) |
| = | 110.56 kN-m |
| Pu/Puz | = | 0.810 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.810, | an | = | 2.000 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((72.83/146.86)2.000) + ((66.97/110.56)2.000) |
| = 0.613 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 32 = 512 mm
iii) 300 mm
Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties 8 @ 300 mm c/c
SUMMARY :
CG16 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 12 nos. (9651 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
Design of CG17 :
General Design Parameters :
Below Floor5 - Default Level at 19.500 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3450 mm |
From analysis results,loads on column
| Axial load,(P) | = | 469.48 kN |
| MomentX,(Mx) | = | 62.65 kN-m |
| MomentY,(My) | = | 31.76 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C25 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 10.282 kN.m |
| My_MinEccen | = | 9.390 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 62.649 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 31.757 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 62.65/469.48 = 133 mm |
| Actual eccenY | = | My / P = 31.76/469.48 = 68 mm |
| eccenXMin | = | (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(133,22) = 133 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(68,20) = 68 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(133 mm) > 0.05 x 300(15 mm)
and eccenY(68 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00278 | 351.94 | 8.92 | 343.02 | 77.59 | 179 | 13.89 |
| 2 | 226 | 0.00137 | 273.45 | 8.03 | 265.42 | 60.04 | 90 | 5.37 |
| 3 | 226 | -0.00004 | -8.30 | 0.00 | -8.30 | -1.88 | 0 | 0.00 |
| 4 | 226 | -0.00145 | -289.67 | 0.00 | -289.67 | -65.52 | -90 | 5.86 |
| 5 | 226 | -0.00286 | -352.67 | 0.00 | -352.67 | -79.77 | -179 | 14.28 |
| Total | -9.54 | | 39.40 |
| xux | = | 222 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 222/450 = 0.178 |
| Puc | = | 0.178 x 20.00 x 300 x 450 |
| = | 480.30 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 480.30 + (-9.54) | = | 470.76 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.D - C2.D) |
| C2 | = | 0.416 ku = 0.416 x 222/450 |
| = | 0.206 |
| Muc | = | 480.30 x (0.5 x 450 - 0.206 x 450) = 63.64 kN-m |
| Mux1 | = | Muc + Mus(Total) |
| | = | 63.64 + (39.40) |
| = | 103.04 kN-m |
For Y axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 565 | 0.00241 | 342.79 | 8.92 | 333.87 | 188.80 | 104 | 19.63 |
| 2 | 565 | -0.00252 | -345.65 | 0.00 | -345.65 | -195.46 | -104 | 20.33 |
| Total | -6.66 | | 39.96 |
| xuy | = | 148 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 148/300 = 0.177 |
| Puc | = | 0.177 x 20.00 x 450 x 300 |
| = | 478.41 kN |
| Puy1 | = | Puc + Pus(Total) |
| = | 478.41 + (-6.66) | = | 471.74 kN > P ,hence O.K. | |
| Muc | = | Puc.(0.5.B - C2.B) |
| C2 | = | 0.416 ku = 0.416 x 148/300 |
| = | 0.205 |
| Muc | = | 478.41 x (0.5 x 300 - 0.205 x 300) = 42.37 kN-m |
| Muy1 | = | Muc + Mus(Total) |
| | = | 42.37 + (39.96) |
| = | 82.34 kN-m |
| Pu/Puz | = | 0.302 | | |
| For Pu/Puz | <= | 0.2, | an | = | 1.0 |
| For Pu/Puz | >= | 0.8, | an | = | 2.0 |
| For Pu/Puz | = | 0.302, | an | = | 1.169 |
| ((Mux / Mux1)an) + ((Muy/Muy1)an) |
| = ((62.65/103.04)1.169) + ((31.76/82.34)1.169) |
| = 0.887 < 1.0,hence O.K. |
Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
Provide 8 mm diameter for ties.Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
not more than the least of the following distances:
i) the least lateral dimension of the compression member = 300 mm
ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
= 16 x 12 = 192 mm
iii) 300 mm
Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties 8 @ 190 mm c/c
SUMMARY :
CG17 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth. |
General Design Parameters :
Below Floor4 - Default Level at 15.600 m| Column size, (B x D) | | = | 300 x 450 mm | Column height,(L) | | = | 3450 mm |
From analysis results,loads on column
| Axial load,(P) | = | 936.89 kN |
| MomentX,(Mx) | = | 38.97 kN-m |
| MomentY,(My) | = | 29.31 kN-m |
| fck | = | 20.00 N/mm˛ |
| fy | = | 415.00 N/mm˛ |
| Load combination | = | 1.50 DL + 1.50 LL |
| Column Name | Orientation Angle |
|---|
| C25 | 0.00 |
Check For Slenderness :| Slenderness RatioX | = | ( Lo x Effective Length FactorX ) / Depth |
| = | ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00, |
column is not slender in this direction.| Slenderness RatioY | = | ( Lo x Effective Length FactorY ) / Width |
| = | ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00, |
column is not slender in this direction.| Mx_MinEccen | = | 20.518 kN.m |
| My_MinEccen | = | 18.738 kN.m |
| Mx | = | max(Mx,Mx_MinEccen) + MuaddX | = | 38.969 kN.m |
| My | = | max(My,My_MinEccen) + MuaddY | = | 29.313 kN.m |
Calculation Of Eccentricities :
As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.| Actual eccenX | = | Mx / P = 38.97/936.89 = 42 mm |
| Actual eccenY | = | My / P = 29.31/936.89 = 31 mm |
| eccenXMin | = | (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm |
| eccenXMin | = | max(22,20)= 22 mm |
| eccenYMin | = | (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm |
| eccenYMin | = | max(17,20) = 22 mm |
| eccenX | = | max(Actual eccenX,eccenXMin) = max(42,22) = 42 mm |
| eccenY | = | max(Actual eccenY,eccenYMin) = max(31,20) = 31 mm |
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(42 mm) > 0.05 x 300(15 mm)
and eccenY(31 mm) > 0.05 x 450(23 mm)
Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :
For X axis :Row No.
(i) | Asi
(mm˛) | ei
| fsi
(N/mm˛) | fci
(N/mm˛) | (fsi - fci)
(N/mm˛) | Pusi
(kN) | xi
(mm) | Musi
(kN-m) |
|---|
| 1 | 226 | 0.00304 | 354.21 | 8.92 | 345.29 | 78.10 | 179 | 13.98 |
| 2 | 226 | 0.00213 | 332.53 | 8.92 | 323.61 | 73.20 | 90 | 6.55 |
| 3 | 226 | 0.00123 | 245.18 | 7.58 | 237.59 | 53.74 | 0 | 0.00 |
| 4 | 226 | 0.00032 | 64.26 | 2.64 | 61.62 | 13.94 | -90 | -1.25 |
| 5 | 226 | -0.00058 | -116.66 | 0.00 | -116.66 | -26.39 | -179 | 4.72 |
| Total | 192.60 | | 24.01 |
| xux | = | 346 mm Puc = C1.fck.B.D |
| C1 | = | 0.36 ku = 0.36 x 346/450 = 0.277 |
| Puc | = | 0.277 x 20.00 x 300 x 450 |
| = | 747.98 kN |
| Pux1 | = | Puc + Pus(Total) |
| = | 747.98 + (192.60) | = | | |